Projectile

Question 1

A stone is thrown horizontally with a velocity of 30 ms⁻¹ from the top of a building 45 m high. Calculate the time of flight.

Answer 1

3 seconds.

Question 2

A stone is thrown horizontally with a velocity of 30 ms⁻¹ from the top of a building 45 m high. Calculate the horizontal distance from the foot of the building to where the stone hits the ground.

Answer 2

90 m.

Question 3

A stone is thrown horizontally with a velocity of 30 ms⁻¹ from the top of a building 45 m high. Calculate the velocity with which the stone hits the ground.

Answer 3

39 ms⁻¹.

Question 4

A bullet is fired horizontally with a velocity of 200 ms⁻¹ from a gun placed at a height of 180 m above a horizontal ground. Calculate the time it takes to reach the ground.

Answer 4

6 seconds.

Question 5

A bullet is fired horizontally with a velocity of 200 ms⁻¹ from a gun placed at a height of 180 m above a horizontal ground. Calculate the range.

Answer 5

1200 m.

Question 6

A bullet is fired horizontally with a velocity of 200 ms⁻¹ from a gun placed at a height of 180 m above a horizontal ground. Calculate the velocity with which it strikes the ground.

Answer 6

208.81 ms⁻¹.

Question 7

A ball is thrown from the ground with a velocity of 20 ms⁻¹ at an angle of 30° to the horizontal. Calculate the time of flight.

Answer 7

2 seconds.

Question 8

A ball is thrown from the ground with a velocity of 20 ms⁻¹ at an angle of 30° to the horizontal. Calculate the maximum height reached.

Answer 8

5 m.

Question 9

A ball is thrown from the ground with a velocity of 20 ms⁻¹ at an angle of 30° to the horizontal. Calculate the range.

Answer 9

34.64 m.

Question 10

A projectile is fired with an initial velocity of 100 ms⁻¹ at an angle of 60° with the horizontal. Calculate the time of flight.

Answer 10

17.32 seconds.

Question 11

A projectile is fired with an initial velocity of 100 ms⁻¹ at an angle of 60° with the horizontal. Calculate the maximum height reached.

Answer 11

375 m.

Question 12

A projectile is fired with an initial velocity of 100 ms⁻¹ at an angle of 60° with the horizontal. Calculate the range.

Answer 12

866.03 m.

Question 13

A projectile is fired with an initial velocity of 50 ms⁻¹ at an angle of 30° to the horizontal. Calculate the maximum range.

Answer 13

250 m.

Question 14

What is a projectile?

Answer 14

Any object that is given an initial velocity and then follows a path determined by the effect of gravity and air resistance alone.

Question 15

What is projectile motion?

Answer 15

The motion of a projectile.

Question 16

Give examples of projectiles.

Answer 16

A ball thrown into the air a bullet fired from a gun an object dropped from an aeroplane.

Question 17

What force acts on a projectile to make it follow a curved path?

Answer 17

The force of gravity.

Question 18

What are the two types of motion a projectile undergoes?

Answer 18

Horizontal motion and vertical motion.

Question 19

Is the horizontal motion of a projectile uniform or accelerated?

Answer 19

Uniform motion.

Question 20

Is there any acceleration in the horizontal direction?

Answer 20

No.

Question 21

What is the horizontal component of the velocity (Vx) throughout the motion?

Answer 21

Constant.

Question 22

What is the vertical component of the velocity (Vy) affected by?

Answer 22

Gravity.

Question 23

What is the vertical component of the velocity (Vy) at the maximum height?

Answer 23

Zero.

Question 24

What is the shape of the path of a projectile?

Answer 24

Parabolic.

Question 25

What is the path of a projectile in the absence of air resistance?

Answer 25

A parabola.

Question 26

What is the angle of projection (θ)?

Answer 26

The angle at which the projectile is launched.

Question 27

What are the initial horizontal and vertical components of the velocity (u)?

Answer 27

ux = u cos θ uy = u sin θ.

Question 28

What is the horizontal distance (x) covered by the projectile at any time t?

Answer 28

x = ux * t = u cos θ * t.

Question 29

What is the vertical distance (y) covered by the projectile at any time t?

Answer 29

y = uy * t - 1/2 * g * t² = u sin θ * t - 1/2 * g * t².

Question 30

What happens when an object like a stone is thrown upwards?

Answer 30

It rises to a particular height and then falls back to the ground.

Question 31

What causes the retardation of the object as it is thrown up?

Answer 31

The gravitational attraction of the earth.

Question 32

What is the acceleration due to gravity?

Answer 32

9.8 ms⁻².

Question 33

What is the value of g?

Answer 33

9.8 ms⁻².

Question 34

Is the acceleration of free fall due to gravity (g) a scalar or vector quantity?

Answer 34

Vector quantity.

Question 35

Is the value of g constant all over the earth's surface?

Answer 35

No.

Question 36

Why does the value of g vary on the earth's surface?

Answer 36

Because the earth is not a perfect sphere and it rotates about its polar axis.

Question 37

What is the acceleration of the stone when moving upwards?

Answer 37

-g.

Question 38

What is the acceleration of the stone when moving downwards?

Answer 38

+g.

Question 39

How are the equations of motion for a body moving under gravity obtained?

Answer 39

By replacing s and a in the standard equations with h and g.

Question 40

What do h and g represent?

Answer 40

h = height of the object above the ground g = acceleration due to gravity.

Question 41

What are the equations of motion under gravity?

Answer 41

v = u ± gt h = ut ± 1/2gt² v² = u² ± 2gh.

Question 42

When is the negative sign used in the equations of motion under gravity?

Answer 42

When the motion is upwards.

Question 43

When is the positive sign used in the equations of motion under gravity?

Answer 43

When the motion is downwards.

Question 44

What is the initial velocity (u) of a body released from a height above the ground?

Answer 44

Zero.

Question 45

What is the velocity (v) of a body thrown upwards at its highest point?

Answer 45

Zero.

Question 46

When does a body have its maximum velocity?

Answer 46

When it hits the ground again.

Question 47

A ball is released from a height above the ground. Find its velocity after 5 seconds. (Take g as 10 ms⁻²).

Answer 47

50 ms⁻¹.

Question 48

A cricket ball is thrown vertically upwards with an initial velocity of 40 ms⁻¹. Find its velocity after 3 secs.

Answer 48

10 ms⁻¹.

Question 49

A cricket ball is thrown vertically upwards with an initial velocity of 40 ms⁻¹. Find the maximum height attained and the time taken to reach it.

Answer 49

Maximum height = 80 m time taken = 4 secs.

Question 50

A cricket ball is thrown vertically upwards with an initial velocity of 40 ms⁻¹. Find the total time taken for the ball to return to the ground again. Neglect air resistance.

Answer 50

Total time = 8 secs.

Question 51

What are the three equations of motion for a body traveling along a straight line with uniform acceleration?

Answer 51

v = u + at s = ut + 1/2at² v² = u² + 2as.

Question 52

Define the variables in the equations of motion.

Answer 52

u = initial velocity v = final velocity t = time s = distance a = uniform acceleration.

Question 53

What is another way to express distance (s)?

Answer 53

s = average velocity x t s = 1/2(u + v)t.

Question 54

A bus traveling at 60 kmhr⁻¹ accelerates uniformly at 5 ms⁻². What is its velocity after 2 minutes?

Answer 54

616.7 ms⁻¹ or 2220.12 kmhr⁻¹.

Question 55

When the brakes are applied to a moving car traveling at 60 kmhr⁻¹ it decelerates at a uniform rate of 5 ms⁻². Calculate the time taken to reach a velocity of 36 kmhr⁻¹.

Answer 55

1.33 seconds.

Question 56

A car accelerates uniformly at a rate of 10 ms⁻² from an initial velocity of 36 kmhr⁻¹ for 30 seconds. Find the distance covered during this period.

Answer 56

4.8 km.

Question 57

A body moving with an initial velocity of 30 ms⁻¹ accelerates uniformly at a rate of 10 ms⁻² until it attains a velocity of 50 ms⁻¹. What is the distance covered during this period?

Answer 57

80 m.

Question 58

A bus moves from rest with a uniform acceleration of 2 ms⁻² for the first 10 s then accelerates at a uniform rate of 1 ms⁻² for another 15 s. It continues at a constant speed for 70 s and finally comes to rest in 20 s by uniform deceleration. Draw the velocity-time graph of the motion.

Answer 58

The velocity-time graph is shown in Fig. 2.6.

Question 59

What is the total distance traveled by the bus?

Answer 59

3.3125 km.

Question 60

What is the average speed for the whole journey?

Answer 60

28.8 ms⁻¹.

Question 61

What is the average retardation as the body is brought to rest?

Answer 61

1.75 ms⁻².

Question 62

What is the maximum speed attained during the motion?

Answer 62

35 ms⁻¹.

Question 63

An aeroplane initially at rest undergoes a constant acceleration of 2.5ms⁻² down the runway for 30s before it lifts off. How far does it travel down the runway before taking off?

Answer 63

1125m.

Question 64

A bus driver moving at a velocity of 100km/hr suddenly sees a goat crossing the highway 49m ahead. He hits hard on his brakes to get a maximum retardation of 8.0m/s². How far will he go before stopping? Can he avoid hitting the goat?

Answer 64

The bus will travel 48.23m before stopping. He will just avoid hitting the goat.

Question 65

How is the total distance travelled by the object during the time interval OC calculated?

Answer 65

By knowing the distances travelled during the time intervals OE ED and DC.

Question 66

How is the distance S₁ covered during the interval OE calculated?

Answer 66

S₁ = average velocity x time = 1/2 * OE * AE = area of triangle OAE.

Question 67

During which time interval does the object move with uniform velocity?

Answer 67

ED.

Question 68

How is the distance S₂ covered during the time interval ED calculated?

Answer 68

S₂ = AE * ED = area of rectangle ABDE.

Question 69

How is the distance S₃ covered during the time interval DC calculated?

Answer 69

S₃ = 1/2 * DC * BD = area of triangle BDC.

Question 70

How is the total distance covered by the object during the time interval OC calculated?

Answer 70

S = S₁ + S₂ + S₃ = sum of the areas of triangle OAE rectangle ABDE and triangle BDC.

Question 71

What is another way to describe the total distance covered?

Answer 71

The area under the velocity-time graph or the area between the velocity-time graph and the time axis.

Question 72

How can the total distance also be obtained?

Answer 72

By finding the area of the trapezium OABC.

Question 73

Give the formula for calculating the total distance S using the trapezium method.

Answer 73

S = 1/2 * (AB + OC) * AE.

Question 74

In the example of a ball thrown vertically upwards how does the velocity change as the ball moves up?

Answer 74

The velocity decreases uniformly from OP to zero.

Question 75

At what point is the ball's velocity zero?

Answer 75

At its highest level (point A).

Question 76

What happens to the velocity as the ball returns to the ground level?

Answer 76

It increases uniformly.

Question 77

What type of motion does the ball undergo as it moves upwards?

Answer 77

Deceleration or retardation.

Question 78

What type of motion does the ball undergo as it returns to the ground?

Answer 78

Acceleration.

Question 79

In the example of a lift moving from one floor to another floor of a tall building how does the velocity change?

Answer 79

The velocity increases to a maximum value and then decreases to zero.

Question 80

Where is the lift at rest?

Answer 80

At one floor and at the other floor.

Question 81

What is obtained when these values are plotted on a graph with velocity on the vertical axis and time on the horizontal axis?

Answer 81

A velocity-time graph.

Question 82

What type of motion is depicted along OA on the graph?

Answer 82

Uniform acceleration.

Question 83

Why is the motion along OA considered uniform acceleration?

Answer 83

Because the velocity is increasing by equal amounts in equal intervals of time.

Question 84

How is acceleration along OA calculated?

Answer 84

change in velocity / change in time.

Question 85

What is the acceleration along OA in Fig. 2.1?

Answer 85

2 ms⁻².

Question 86

What type of velocity is observed along AB on the graph?

Answer 86

Uniform or constant velocity.

Question 87

What is the acceleration along AB?

Answer 87

Zero.

Question 88

Define uniform acceleration.

Answer 88

The motion of an object whose velocity increases by equal amounts in equal intervals of time.

Question 89

Give an example of uniform acceleration.

Answer 89

The motion of a body falling freely under gravity.

Question 90

Define uniform or constant velocity.

Answer 90

The velocity of a moving body where equal distances are covered in equal time intervals.

Question 91

What happens to the velocity along BC on the graph?

Answer 91

It decreases uniformly from 10 to 0.

Question 92

What is the body said to experience along BC?

Answer 92

Retardation or deceleration.

Question 93

Define uniform deceleration.

Answer 93

Velocity decreases by equal amounts in equal times.

Question 94

What is the deceleration along BC in Fig. 2.1?

Answer 94

2ms⁻².

Question 95

What is another term for retardation or deceleration?

Answer 95

Negative acceleration.

Question 96

Is acceleration a scalar or vector quantity?

Answer 96

Vector quantity.

Question 97

What is observed in a velocity-time graph for variable acceleration?

Answer 97

Irregular velocity variation.

Question 98

How is acceleration determined at a point C on a variable acceleration graph?

Answer 98

By the gradient of the tangent to the graph at that point.

Question 99

What is the relationship between acceleration and the velocity-time graph?

Answer 99

Acceleration = gradient of velocity-time graph.