Question

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?

Answer 1

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