# Question #b5ed9

##### 1 Answer
Mar 16, 2017

Question

A baseball leaves a pitcher's hand horizontally at a speed of 151 km/h. The distance to the batter is 18.3 m. (Ignore the effect of air resistance.) (a) How long does the ball take to travel the first half of that distance? (b) The second half? (c) How far does the ball fall freely during the first half? (d) During the second half?

Answer
As there exists no air resistance the base ball will continue its motion in horizontal direction with initial velocity given by the pitcher.

$v = 151 \text{km/hr"or(151xx10^3)/3600"m/s}$

The distance of batter from pitcher is $d = 18.3 m$

The ball will take same time to cover first or second half of this distance and this time will be
$t = \frac{\frac{d}{2}}{v} = \left(\frac{18.3}{2}\right) \cdot \frac{3600}{151 \times {10}^{3}} s \approx 0.218 s$

Let the ball falls ${h}_{1} m$ freely during first half of its journey towards batter

So by equation of kinematics we have

${h}_{1} = \frac{1}{2} \cdot g \cdot {t}^{2} = \frac{1}{2} \times 9.8 \times {\left(0.218\right)}^{2} = 0.233 m$

So the height of free fall in second half will be

${h}_{2} = \frac{1}{2} \cdot g \cdot {\left(2 t\right)}^{2} - \frac{1}{2} \cdot g \cdot {t}^{2}$
$= \frac{1}{2} \cdot g \cdot 3 {t}^{2} = 3 \cdot {h}_{1} = 0.799 m$