Projectile motion

Equations of motion

Questions

A ball is thrown upwards at 30 m•s-1 from the top of a high cliff. Calculate the:a. ball’s maximum height it reaches above the point it

leaves the thrower’s hand.b. ball’s position 4 s after leaving the thrower’s hand.c. ball’s velocity 4 s after leaving the thrower’s hand.d. ball’s position 6,5 s after leaving the thrower’s hand.e. ball’s position 3 s after being at its highest point.f. time(s) when the ball is 5 m above the thrower’s

hand.

Maximum height

a. Calculate the maximum height it reaches above the point it leaves the

thrower’s hand.

+ 30 m•s-1

0 m•s-1

∆ 𝑦=?m•s-2

a. Maximum height

vf 2 = vi 2 + 2ay

(0)2 = (30)2 + 2(-9,8)y0 = 900 – 19,6 y19,6 y = 900y = y = 45,92 my = 45,92 m above the point it leaves the hand

+ 30 m•s-1

0 m•s-1

∆ 𝑦=45,92 mm•s-2

∆ 𝑦=45,92 m

Height at given time 1

b. Calculate its position 4 s after leaving the thrower’s hand.

∆ 𝑡=4  s

∆ 𝑡=4  sm•s-2

∆ 𝑦=?

+ 30 m•s-1

b. Height at 4 s

y = vi t + ½ at2

y = (30)(4) + ½ (-9,8)(4)2

y = 120 + (-78,4)y = 41,6 my = 41,6 m above the point it leaves the hand

∆ 𝑡=4  s∆ 𝑦=41,6 m

+ 30 m•s-1

Velocity at given time

c. Calculate its velocity 4 s after leaving the thrower’s hand.

∆ 𝑡=4  s∆ 𝑦=41,6 m

m•s-2

?

+ 30 m•s-1

b. Velocity at 4 s

vf = vi + at

vf = 30 + (-9,8)(4)

vf = 30 – 39,2

vf = - 9,2 m•s-1

vf = 9,2 m•s-1 downward

= -9,2 m•s-1

+ 30 m•s-1

∆ 𝑡=4  s∆ 𝑦=41,6 m

m•s-2

= -9,2 m•s-1

Height at given time 2

d. Calculate its position 6,5 s after leaving the thrower’s hand.

?

+ 30 m•s-1

∆ 𝑡=6,5   s

∆ 𝑦=?

m•s-2

d. Position at given time 2

y = vi t + ½ at2

y = (30)(6,5) + ½ (-9,8)(6,5)2

y = 195 + (-297,025)y = -12,025 my = 12,03 m below the point it leaves the hand

∆ 𝑦=12,03  m

+ 30 m•s-1

∆ 𝑡=6,5   s

∆ 𝑦=12,03  m

m•s-2

Height at given time 3

e. Calculate its position 3 s after being at its highest point.

∆ 𝑡=3  s

∆ 𝑡=3  s

0 m•s-1

m•s-2

∆ 𝑦=?

e. Position at given time 3

y = vi t + ½ at2

y = (0)(3) + ½ (-9,8)(3)2

y = 0 + (-44,1)y = -44,1 my = -44,1 m below the highest point(= 1,82 m above the point it leaves the hand)

∆ 𝑦=  44,1 m

∆ 𝑡=3  s

0 m•s-1

m•s-2

∆ 𝑦=  44,1 m