A ball is thrown at a speed of 25 m s^-1 at an angle of 53.10 (measured upward f

Question

A ball is thrown at a speed of 25 m s^-1 at an angle of 53.10 (measured upward from the horizontal) from the top of a 25 m tall cliff. a. What are the initial components of the velocity vector? b. What is the maximum height reached by the ball, and when does it reach it? What is the distance from the release point at this time? c. What are the components of the velocity when the ball hits the ground? d. What is the ball's speed when it hits the ground? e. How far from the base of the cliff does the ball land?

Explanation / Answer

The motion is a projectile motion

a)

Initial components of velocity are

vx = vcostheta = 25*cos 53.1

vx = 15.01 m/s

Vertical component, vy = v sintheta

vy = 25 sin 53.1

vy = 19.9 m/s

b)

Maximum height reached by ball is

H = v^2 sintheta^2/2g

H = 25^2*sin53.1^2/2*9.8

H = 20.39 m from top of cliff

Time taken is t = vsintheta/g

t = 25 sin 53.1 /9.8

t = 2.04 seconds

Distance from release point is 20.39 m

c)

When ball hits ground,

Vx remains same = 15.01 m/s

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