Question

An object is launched at an angle of 24 degrees above the horizontal at a velocity of 33.7 m/s from the top of a building which is 110.8 meters above the ground. When the object lands on the ground, what is the horizontal displacement from the base of the building in meters?

Hint: First find the final velocity of the object in the y direction when it hits the ground using the given information above. Then use this final velocity in the y direction to find the time it takes to achieve this velocity in the y direction using a simple projectile motion equation. With this time it should be straight forward to find the horizontal displacement.

Answer 1

Vo = 33.7m/s[24o]
Xo = 33.7*Cos24 = 28 m/s.
Yo = 33.7*Sin24 = 13.7 m/s.

Tr = -Yo/g = -13.7/-9.8 = 1.40 s. = Rise time.

h max = -Yo^2/2g = -(13.7^2)/-19.6 = 9.58 m. Above the bldg.

0.5g*t^2 = 9.58 + 110.8 = 120.4 m
4.9*t^2 = 120.4
t^2 = 24.57
Tf = 4.96 s. = Fall time.

Dx = Xo*(Tr+Tf) = 28 * (1.40+4.96) = 176.1 m.