Question

a soccer ball is kicked horizontally off a bridge with a height of 36m. The ball travels 25m horizontally before it hits the pavement below. What was the soccer ball’s speed when it was first kicked?

Answer 1

The initial velocity at which the ball is kicked is 9.22 m/s

Step-by-step explanation:

Given

bridge height, h = 36 m

Horizontal distance at which ball falls,
`d_(x)` = 25 m

Firstly, we have to find the time taken by ball to reach vertically. It is calculated using formula,

d =
`V_(i)` t +
`(1)/(2)`
`at^(2)`

where d is vertical distance, d = h and a is acceleration due to gravity, a = g = 9.8 m/
`s^(2)`. Also initially ball is at rest, hence
`V_(i)` = 0. Substituting the values in the formula, we get

-36 = 0 +
`(1)/(2)` (-
`9.8t^(2)`) (negative sign is because the object moves downward)

`t^(2)` = 7.347

t =
`√(7.347)`

= 2.711 s

To find: initial velocity

We know that horizontal velocity is derived from horizontal distance and time:

`V_(x)` =
`(d_(x) )/(t)`

=
`(25 m)/(2.711 s)`

= 9.22 m/s