Calculate the maximum height above the roof reached by the r

Calculate the maximum height above the roof reached by the rock. A man stands on the roof of a 13.0 m -tall building and throws a rock with a velocity of magnitude 30.0 m/s at an angle of 38.0 n above the horizontal. You can ignore air resistance. Calculate the magnitude of the velocity of the rock just before it strikes the ground. Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

Solution

A. Initial vertical component of velocity,

uy = 30 sin38 = 18.47 m/s

a = -9.8 m/s^2

using vf^2 - vi^2 = 2ad

foe max height vf = 0

0^2 - 18.47^2 = 2(-9.8)(H)

H = 17.4 m

b) vy^2 - uy^2 = 2ad

vy^2 - 18.47^2 = 2(-9.8)(-13)

vy = 24.41 m/s

and vx = ux = 30cos38 = 23.64 m/s

v = sqrt(vx^2 + vy^2) = 33.98 m/s

c) vy = uy + at

- 24.41 = 18.47 - 9.8t

t = 4.38 sec

d = vx *t = 23.64 x 4.38 = 103.44 m