Question

1. A ball is launched horizontally at 40 m/s while 6.0 m above a level field . How far will the ball move horizontally before striking the ground ?

Answer 1

Answer:

44 m

Explanation:

Given that,

Horizontal velocity of the ball, u = 40 m/s

It is 6 m above the level field.

We need to find the distance covered by the ball when move horizontally before striking the ground. Let it is d.

Firstly, we will find time taken for the ball to hit the ground. Using second equation of motion as follows :

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

Put u = 0 and a = g

$s=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2s}{g}} \\\\t=\sqrt{\dfrac{2\times 6}{9.8}} \\\\t=1.1\ s$

No finding the horizontal distance as follows :

d = ut

d = 40 m/s × 1.1 s

d = 44 m

So, the ball will move 44 m horizontally before striking the ground.