A baseball is hit almost straight up into the air with a speed of$\mathbf{25}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}$. Estimate (a) how high it goes and (b) how long it is in the air. (c) What factors make this an estimate?

The rate of change of velocity that is produced due to the gravitational pull of the earth is called acceleration due to gravity.

The value of acceleration due to gravity is generally taken to be$g=9.8{\mathrm{ms}}^{-2}$, in the downward direction.

The initial speed of the ball,$u=25{\mathrm{ms}}^{-1}$.

Assume that the maximum height achieved by the ball is h, and the ball stays for $t_{\circ }$time in the air.

The acceleration due to gravity is $\mathrm{g}=-9.8{\mathrm{ms}}^{-2}$. (Taking the upward direction as positive)

When the ball reaches its maximum height, its velocity will become zero. Thus, using the third equation of motion,

$0^2-u^2=2\mathrm{g}h$.

Substituting the values in the above equation,

$0-{25}^2=2\times \left(-9.8\right)\times h$.

Solving for the value of h,

$h=31.89\mathrm{m}$.

Thus, the maximum height achieved by the ball is 31.89 m.

From the second equation of motion, you can write the displacement of the ball at time t as

$s=u\times t+\frac{1}{2}\mathrm{g}{t}^2$.

Here, as the initial and final positions are the same, the displacement of the ball is zero. Thus,

$0=25\times t_{\circ }+\frac{1}{2}\times \left(-9.8\right)\times t_{\circ }^2$.

Solving the above quadratic equation and neglecting root t = 0,

$t_{\circ }=5.1\mathrm{s}$.

Thus, the total time for which the ball was in the air is 5.1 s.

• The value of acceleration due to gravity has been taken as \[9.8{\rm{ m }}{{\rm{s}}^{ - 2}}\], but it can affect the estimation as it varies from place to place.
• You have assumed that wind has no effect, but wind can cause drag and even force the ball to change its trajectory. That is why wind resistance can also affect the estimation.