Question

A 0.49kg rock is projected from the edge of the top of a building with an initial velocity of 8.36 m/s at an angle 60◦ above the horizontal.Due to gravity, the rock strikes the ground at a horizontal distance of 10.7 m from the base of the building.How tall is the building?

The acceleration of gravity is
9.8 m/s2 .

Answer 1

A 0.49kg rock is projected from the edge of the top of a building with an initial velocity of 8.36 m/s at an angle 600; above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 10.7 m from the base of the building. How tall is the building?

vy0 = v0 * sin60° = 8.36 * √3/2 = 7.24 m/s
vx0 = v0 * cos60° = 8.36 * 1/2 = 4.18 m/s

Total time of flight:

t = d ÷ vx0 = 10.7m ÷ 4.18 m/s = 2.56 s

Time of flight over the top of the building level:

t′ = 2t_peak = 2 * vy0 ÷ g = 2 * 7.24 m/s ÷ 9.8 m/s² = 1.48 s

Hence, height of the building is:

h = vy0 * (t - t′) + g * (t - t′)² ÷ 2 = 7.24 m/s * (2.56 s - 1.48 s) + 9.8 m/s² * (2.56 s - 1.48 s)² ÷ 2 = 13.53 m