A batsman deflects a ball by an angle of `45^(@)` without changing its initial speed which is equal to `54 km/h` . What is the impulse imparted to the ball ? ( Mass of the ball is 0.15 kg)

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 decending 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`.