Question

A steel projectile is shot horizontally at 20m/s from the top of a 40m tower. How long does it take to hit the ground? How far from the base of the tower does the projectile hit the ground

Answer 1

1) 2.86 s

To find the time of flight, we have to analyze the vertical motion of the stone, which is a uniformly accelerated motion (free fall) with constant acceleration (acceleration of gravity).

The vertical displacement at time t is given by

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

where , taking downward as positive direction

$u$ is the initial vertical velocity  (which is zero, since the projectile is shot horizontally)

t is the time

$g=9.8 m/s^2$ is the acceleration of gravity

In this problem we have

s = 40 m (vertical displacement is the height of the tower)

Solving for t, we find the time of flight: of the projectile:

$s=\frac{1}{2}gt^2\\t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(40)}{9.8}}=2.86 s$

So, the projectile takes 2.86 s to hit the ground.

2) 57.2 m

The horizontal motion of the projectile is a uniform motion: in fact there are no forces acting on the projectile along the horizontal direction, so the horizontal acceleration is zero, and so the horizontal velocity is constant.

The initial horizontal velocity of the projectile is

$v_x = 20 m/s$

Therefore, to find the horizontal distance covered by the projectile, we can just use the equation

$d=v_x t$

And substitute the time of flight, which is the time at which the projectile hits the ground:

t = 2.86 s

So, we get:

$d=(20)(2.86)=57.2 m$