Question

# A ball is thrown toward a cliff of height h with a speed of 30m/s and...

A ball is thrown toward a cliff of height h with a speed of 30m/s and an angle of 60 degrees above horizontal. it lands on the edge of the cliff 4 seconds later.

A) How high is the cliff?

B)What is the maximum height of the ball?

C)What is the ball's impact speed?

At first we have to find the horizotal and vertical components of the velocity

Velocity in x direction:
v = 30 * cos 60 = 15 m/s
Distance traveled (x )= 30 * t * cos 60 = 30 * 4 * cos 60 = 60 meters

Velocity along y direction:
a = -g
v = -g * t + 30 * sin 60
= -9.8 * 4 + 30 * sin 60
= -13.219 m/s

Now,

a) height of cliff =

Finding distance traveled along y axis
y = -(1/2) * g * t^2 + 30 * t * sin 60
= -(1/2) * 9.8 * 4^2 + 30 * 4 * sin 60
= -78.4 + 103.923
= 25.523 m
The cliff is 25.5 meters high.

b) maximum height reached by ball = v^2sin^2(60)/2*g

= 30^2*sin^2(60)/2*9.8

= 34 .43 m

C) using the formula

vf^2 = vx^2 + vy^2
= (15)^2 + (-13.219)^2
= 399.748
vf = 19.994 m

Therefore ball impact velocity = 20m/s