A batsman hits a ball at an angle of `306^@` with an initial speed of `30 ms^-1`. Assuming that the ball travels in a verticle 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.

Explanation

Here `theta = 30^@ , u = 30 m 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_H = (T)/(2) = (u sin theta)/(g) = (30)/(10) xx sin 30^@ = 1.5 s`
(b) The maximum height reached is
`(u^2 sin^2 theta)/(2 g) = ((30)^2 xx (sin 30^@)^2)/(2 g) = (900)/(2 xx 10 xx 4) = 11.25 m`
( c) Horizontal range `= (u^2 sin 2 theta)/(g)`
`=((30)^2 sin 2 (30^@))/(10)`
`= (900sqrt(3))/(20)m = 45 sqrt(3) m`
(d) The time for which the ball is in air is same as its time of flight, i.,e.,
`(2 u sin theta)/(g) = (2 xx 30 xx sin 30^@)/(10) = 3 s`.