Question

A small, dense ball is launched from ground level at an angle of 50° above the horizontal. The ballâs initial speed is 22 m/s, and it lands on a hard, level surface at the same height from which it was launched. The ball then bounces and reaches a height of 75% its peak height it achieved at launch. Assume air resistance is negligible.(a) Find the maximum height reached by the ball during its first parabolic arc. (b) Find the distance between the launch point to where the ball lands the first time. (c) Find the distance between the launch point to where the ball lands the second time?

Answer 1

a) Hmax = 10.86 m

b) Rx = 48.64 m

c) Rx¹ = 36.48m

Step-by-step explanation:

Given that Ф = 50°

v₀ = 22 m/s

a)

hmax = v₀²sin²Ф / 2g

hmax = 22² × sin²50 / 2 × 9.8

hmax = 14.49 m

Hmax = 75% of hmax

Hmax = 0.75 × 14.49

Hmax = 10.86 m

the maximum height reached by the ball during its first parabolic arc is 10.86 m

b)

Rx = v₀²sin2Ф / g

Rx = 22² × sin 100 / 9.8

Rx = 48.64 m

the distance between the launch point to where the ball lands the first time is 48.64 m

c)

Rx¹ = 75% 0f Rx

Rx¹ = 0.75 × 48.64

Rx¹ = 36.48m

the distance between the launch point to where the ball lands the second time is 36.48m