Answer to Question #170718 in Atomic and Nuclear Physics for Nema

Question

You throw a ball upwards with an initial velocity of 29 m/s. How high will the ball go before it%92s velocity will be zero and it begins to fall and how much time will it take for the ball to reach that height ?

Answer 1

Let us write the equation of motion for the ball. The initial height is 0, or we may suppose we place the origin of the vertical axis at the initial position of the ball. At the time t the height will be

$h(t) = h_0 + v_0t +\dfrac{-gt^2}{2} = v_0t - \dfrac{gt^2}{2} = 29t - \dfrac{9.8t^2}{2}.$

Here t is measured in seconds.

The dependence of velocity on time is

$v(t) = v_0-gt,$ so the time interval from start to the moment with zero velocity will be

$t=\dfrac{v_0}{g} =\dfrac{29}{9.8} \approx 3\,\text{s}.$

So the height at the moment will be

$h(3) =29\cdot3 - \dfrac{9.8\cdot 3^2}{2} = 42.9\,\text{m}.$