Question

A rock is thrown from the edge of the top of a 51 m tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 74 m from the base of the building 8 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.

Answer 1

Answer:

The speed of projection is 34 m/s.

Explanation:

Height of building, h = 51 m

horizontal distance, d = 74 m

time, t = 8 s

Let the angle is A and the speed is u.

d = u cos A x t

74 = u cos A x 8

u cos A = 9.25 .... (1)

Use second equation of motion

$h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)$

Squaring and adding both the equations

$u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s$