Question

A tiger leaps horizotally from a 15.5 m high rock with a speed of 7.5 m/s. How

Answer 1

The tiger lands 13.4 m from the base of the rock

Explanation:

The motion of the tiger is a projectile motion, therefore it consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

We start by analyzing the vertical motion, to find the time of flight of the tiger. Using the following suvat equation:

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

where, taking downward as positive direction, we have:

s = 15.5 m is the vertical displacement of the tiger (the height of the rock)

$u=0$ is the initial vertical velocity

t is the time of the fall

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

Solving for t, we find :

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(15.5)}{9.8}}=1.78 s$

Therefore the tiger takes 1.78 s to land on the ground.

Now we analyze the horizontal motion: the tiger moves horizontally with a constant velocity of

$v_x = 7.5 m/s$

So, the distance it covers in a time t is given by:

$d=v_x t$

and by substituting t = 1.78 s, we find the horizontal distance covered by the tiger:

$d=(7.5)(1.78)=13.4 m$

Learn more about projectile motion:

brainly.com/question/8751410

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