Question

A cat is shot horizontally out of a cannon that is on top of a 125 meter tall cliff. If the cat lands 286 m away from the base of the cliff, how fast did the cat come out of the cannon

Answer 1

The initial velocity of the cat is 56.6 m/s

Explanation:

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

- a uniform horizontal motion with constant velocity

- a vertical accelerated motion with constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time it takes for the cat to reach the ground. We use the following suvat equation:

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

where we have (choosing downard as positive direction)

s = 125 m is the vertical displacement of the cat

u = 0 is the initial vertical velocity

t is the time taken to reach the ground

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

Solving for t, we find

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(125)}{9.8}}=5.05 s$

Now we can analzye the horizontal motion. Since the motion is uniform and the horizontal velocity is constant, it is given by

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

where

d = 286 m is the horizontal distance travelled

t = 5.05 s is the time taken

Solving the equation,

$v_x = \frac{286}{5.05}=56.6 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

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