Question

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80-m high cliff. The shell strikes the ground 1330 m from the base of the cliff. The drawing is not to scale.

Answer 1

Missing question:

What is the initial speed of the shell?

Answer:

329.2 m/s

Explanation:

The motion of the shell is a projectile motion, consisting of:

- A uniform motion along the horizontal direction, with constant velocity $v_x$

- A uniformly accelerated motion along the vertical direction, due to the force of graviity

We start by analyzing the vertical motion first. The shell falls down due to the effect of gravity, so we can use the following equation:

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

s = 80 m is the vertical displacement of the shell (= the height of the cliff)

u = 0 is the initial vertical velocity

t is the time of flight

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

Solving for t, we find the time of flight:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(80)}{9.8}}=4.04 s$

Now we know that along the horizontal direction, the shell travels a distance of

d = 1330 m

And it takes a time of

t = 4.04 s

We know that the horizontal velocity is constant, so we can find it as:

$v_x = \frac{d}{t}=\frac{1330}{4.04}=329.2 m/s$

And since it is constant during the entire motion, this was also the initial horizontal velocity of the bullet.