Question

1) A tiger leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from

Answer 1

The distance of tiger's leap from the base of rock is 5.58 m

It is a question of two dimensional motion

The time of motion in two dimensional motion is given by:

t= $\sqrt{2y/g}$

where y is the height and g is  the acceleration due to gravity

y is given to be 7.5m and let us assume g to be 9.8 m/s^2

t = $\sqrt{2 * 7.5/9.8}$

 = 1.24s

Using time and speed,

We know that distance is the product of speed and time,

Distance= speed x time

speed is given to be 4.5 m/s

distance from the base of rock = 4.5 x 1.24

                                                    = 5.58m

Hence the distance of tiger's leap from the base of rock is 5.58 m

Disclaimer:

The acceleration due to gravity is assumed to be 9.8 m/s^2

For further reference:

brainly.com/question/11213880?referrer=searchResults

#SPJ9