Question

A batsman hits a ball at an angle of ${30}^o$ with an initial speed of 30 m s${}^{-1}$ Assuming that the ball travels in a vertical plane, calculate
a. The time at which the ball reaches the highest point
b. The maximum height reached
c. The horizontal range of the ball
d. The time for which the ball is in the air.

Answer 1

Here $\theta ={30}^o,u=30m{s}^{-1}$
a. The time taken by the ball to reach the highest point is half the total time of flight. As the time of ascending and descending is same for a projectile without air resistance. the time to reach the highest point T
$t_H=\frac{T}{2}=\frac{usin\theta }{g}=\frac{30}{10}\times sin{30}^o=1.5s$
b. The maximum height reached is
$\frac{u^{2}si{n}^2\theta }{2g}=\frac{\left({30}^2\right)\times (sin{30}^o{)}^2}{2g}=\frac{900}{2\times 10\times 4}=11.25m$
c. Horizontal range = $\frac{u^{2}sin2\theta }{g}$
= $\frac{\left(30{)}^{2}sin2\right({30}^o)}{10}$
= $\frac{900\sqrt{3}}{20}=45\sqrt{3}m$
d. The time for which thc ball is in air is same as its time of eight, i.e.,
$\frac{2usin\theta }{g}=\frac{2\times 30\times \sin {30}^o}{10}=3s$