Week 7 - Uniform motion

Question 1

All projectiles move with a constant acceleration, what is their vertical and horizontal acceleration?

Answer 1

Vertical = -9.81 m/s2 due to gravity
Horizontal = 0 m/s2

Question 2

Equations for uniform motion

Answer 2

𝑣𝑓 = 𝑣𝑖 + 𝑎*𝑡
𝑣𝑓2 = 𝑣𝑖2 + 2𝑎*∆𝑑
∆𝑑 = 𝑣𝑖*𝑡 +1/2𝑎*𝑡2

Question 3

What is the final velocity of a ball kicked if the initial velocity is 40m/s and it lands on the same level surface?

Answer 3

40m/s as vi = vf if they are on the same surface

Question 4

You sling shot a bird with a resultant velocity of 6.0 m/s at a
release angle of 50°. The initial horizontal velocity is 3.9 m/s and
the initial vertical velocity is 4.6 m/s.
What is the maximum vertical height the bird reaches?
Which equation will you choose to solve the answer?

Answer 4

𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆d

Rearrange the equation to get ∆d on its own

𝑣𝑓2 - 𝑣𝑖2 = 2𝑎∆d

(𝑣𝑓2 - 𝑣𝑖2)/(2*𝑎) = ∆d

Question 5

You sling shot the bird with a resultant velocity of 6.0 m/s at a
release angle of 50°. The initial horizontal velocity is 3.9 m/s and
the initial vertical velocity is 4.6 m/s.
How long is the bird in the air?
Which equation will you choose to solve the answer?

Answer 5

𝑣𝑓 = 𝑣𝑖 + 𝑎*t

Rearrange the equation to get t on its own

𝑣𝑓 - 𝑣𝑖 = 𝑎*t

(𝑣𝑓 - 𝑣𝑖)/𝑎 = t

total time = 2*t

Question 6

You sling shot the bird with a resultant velocity of 6.0 m/s at a
release angle of 50°. The initial horizontal velocity is 3.9 m/s and
the initial vertical velocity is 4.6 m/s.
What is the Horizontal Range (horizontal distance)?
Which equation will you choose to solve the answer?

Answer 6

∆𝑑 = 𝑣𝑖𝑡 +1/2𝑎𝑡2

∆𝑑 = (3.90.94) + 1/20*0.94squared

Calculate half of the equation out

∆𝑑 = 3.9*0.94

∆𝑑 = 3.67 m horizontal range