HDE Section 2: Projectiles

Question 1

Horizontal and vertical motion are considered ____________

Answer 1

Horizontal and vertical motion are considered separately

Question 2

What acceleration does all vertical motion have?

Answer 2

Vertical motion will have an acceleration due to gravity (9.81ms-1)

Question 3

Students collect three measurements for the acceleration due to gravity: 12 ms^-2, 8.5 ms^-2 and 9.1 ms-2.
Discuss the precision of these results.
Discuss the accuracy of these results.
Describe how these results could be improved.

Answer 3

They are not precise or accurate

Take more readings

Question 4

Describe the difference between a scalar and a vector.

Answer 4

Scalar does not have a direction

Question 5

A golf ball is launched at 19 ms-1 at an angle of 65 degrees to the horizontal.

a) Sketch this example.
b) Resolve into vertical and horizontal components of velocity.
c) Calculate how long it takes for the golf ball to reach its peak.

Answer 5

a) 17 horizontal 8 vertical
b) cos(65) x 19 = 8
sin(65) x 19 = 17
c)
s =
u = 0
v = 8
a = -9.81
t = ?

(v-u)/a
8/9.81 = 0.86s

Question 6

A ball is thrown with an initial velocity of 20ms^-1 at an angle of 30° to the horizontal. How far will it travel before it lands? (For simplicity, take g as 10 ms^-1)

Answer 6

sin(30) x 20 = 10
cos(30) x 20 = 17.32

Vertical
s = 
u = 0
v = 10
a = -9.81
t = ?

v-u / a 10/9.81 = 1.019
t = 1.019

Horizontal 
s = 
u = 0
v = 17.32
a = 
t = 2.039

(17.32 x 2.039)/2 = 17.6555m

Question 7

A cannonball is fired horizontally from the top of a vertical cliff, 60 m above sea level, with a velocity of 200 ms-1.
How far from the cliff face does the cannonball enter the sea?
What assumption have you made?

Answer 7

Horizontal and vertical components of the velocity are independent. In the vertical plane both coins will accelerate downwards at 9.8 ms^-2 over the same vertical displacement.
a) In the vertical plane: s = ut + 1/2at^2
60 m = 0 + 0.5 x 9.8 ms^-2 x t^2 t = 3.5 s
In the horizontal plane: s = 700 m
b) Air resistance is not taken into account.

Question 8

An arrow is fired at 40o to the horizontal with velocity 40 ms-1. Calculate:

a) The horizontal and vertical components of the velocity.
b) The time taken to hit a target in the same horizontal plane as the release point of the arrow.
c) The horizontal distance of the target from the archer.

Answer 8

a) uh= 40 cos 40 ° ms^-1 = 31 m s-1
b) uv = 40 sin 40 ° ms^-1 = 26 m s-1
c) In the vertical plane, v = u + a t 0= 26 m s-1 - 9.8 m s-2 x t where t = time to reach maximum height = 2.7 s Time in flight = 2 t = 5.4 s d) Range = horizontal velocity x time = 31 m s-1 x 5.4 s = 167 m

Question 9

True or false?

Acceleration is a scalar.
Results that are close to the true value are accurate.
The vertical component of a force at 20o to the vertical is equal to
F sin (20).
v2 = u2 - 2as
N = ms-1
A 6 N force combines with a 4 N force. This resultant force is equal 52 N.
Projectiles moving with a greater starting velocity will always travel higher.

Answer 9

F
T
F
F
F
F