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

9.23 m/s

Explanation:

Let, the initial velocity be ux.

The horizontal velocity remans the same. So, time required, t = 25/ux

For vertical component, we know,

h = uy*t + (1/2)*g*t^2    [ g is positive because ball is falling downward ]

Putting in the values, we get,

36 = 0*(25/ux) + 1/2 * 9.81 * (25/u)^2

36 = 3065.625/u^2

u^2 = 85.15625

u = 9.23

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