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

Answer:

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

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 + $\frac{1}{2}$ $at^{2}$

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 + $\frac{1}{2}$ (-$9.8t^{2}$)   (negative sign is because the object moves downward)

$t^{2}$ = 7.347

t = $\sqrt{7.347}$

 = 2.711 s

To find: initial velocity

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

$V_{x}$ = $\frac{d_{x} }{t}$

     = $\frac{25 m}{2.711 s}$

     = 9.22 m/s