# A soccer ball is kicked from the playing field at a 45° angle. If the ball is in the air for 3 s, what is the maximum height achieved? Hello!

Denote the angle as `alpha,` the initial speed as `V` and the given time as `T.`

I suppose we ignore air resistance. Then the only force acting on the ball is the gravity force, it is directed downwards and gives the acceleration `g = 9.8 m/s^2` to the ball.

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Hello!

Denote the angle as `alpha,` the initial speed as `V` and the given time as `T.`

I suppose we ignore air resistance. Then the only force acting on the ball is the gravity force, it is directed downwards and gives the acceleration `g = 9.8 m/s^2` to the ball.

The vertical component of the velocity uniformly decreases with time `t` from `V sin(alpha)` with the speed `g,` so it is equal to `V sin(alpha) - g t.` The height itself is equal to `H(t) = V sin(alpha) t - (g t^2)/2.` At the time `T` the velocity is zero, i.e. `V sin(alpha)T =(g T^2)/2,` or `V sin(alpha) = (g T)/2.`

The maximum height is reached when the vertical speed becomes zero, i.e. when `V sin(alpha) = g t.` From the above we know that this time is `T/2.`

Finally, the maximum height is

`H(T/2) =Vsin(alpha) T/2 - (g T^2)/8 =(g T^2)/4 -(g T^2)/8 =(g T^2)/8.`

Numerically it is `(9.8*9)/8 approx 11 (m).` This is the answer. Note that it doesn't depend on `alpha.`

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