Question

A rock is thrown from the edge of the top of a 100-ft tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 160 ft from the base of the building 5.0 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:

133.9 ft/s

Explanation:

We can find the horizontal component and the vertical component of the initial velocity by considering the horizontal and vertical motions indipendently.

Along the horizontal direction, the rock travels at constant speed: this means that its horizontal velocity is constant, and it is given by

$u_x = \frac{d}{t}$

where

d = 160 ft is the distance covered

t = 5.0 s is the time taken

Substituting,

$u_x = \frac{160}{5}=32 ft/s$

Along the vertical direction, the rock is in free-fall - so its motion is a uniform accelerated motion with constant acceleration g = -32 ft/s^2 (downward). Therefore, the vertical distance covered is given by the suvat equation:

$s=u_y t + \frac{1}{2}gt^2$

where

s = -100 ft is the vertical displacement

$u_y$ is the initial vertical velocity

Replacing t = 5.0 s and solving the equation for $u_y$, we find:

$u_y = \frac{s}{t}-\frac{1}{2}gt=\frac{100}{5}-\frac{1}{2}(-32)(5)=130 ft/s$

Therefore, the initial speed is just the magnitude of the initial velocity:

$u=\sqrt{u_x^2+u_y^2}=\sqrt{32^2+130^2}=133.9 ft/s$