Question

A ball is thrown toward a cliff of height h with a speed of 25m/s and an angle of 60? above horizontal. It lands on the edge of the cliff 3.5s later.

Part A

How high is the cliff?

Part B

What was the maximum height of the ball?

Part C

What is the ball's impact speed?

Answer 1

Given initial speed, vo = 25 m/s

y-component of initial velocity, voy = vo*sin(60)

= 25*sin(60)

= 21.65 m/s

x-component of initial velocity, vox = vo*cos(60)

= 25*cos(60)

= 12.5 m/s

A) Let H is the height of the cliff

Apply, H = voy*t - 0.5*g*t^2

H = 21.65*3.5 - 0.5*9.8*3.5^2

H = 15.75 m <<<<<<<--------------------------Answer

B) maximum height reached by the ball, Hmax = voy^2/(2*g)

= 21.65^2/(2*9.8)

= 23.91 m <<<<<<<--------------------------Answer

C) y-component of velosity at the impact, vy = voy - g*t

= 21.65 - 9.8*3.5

= -12.65

x-component of velosity at the impact,vx = vox

= 12.5 m/s

impact speed, v = sqrt(vx^2 + vy^2)

= sqrt(12.5^2 + 12.65^2)

= 17.78 m/s <<<<<<<--------------------------Answer