Two stones `A and B` are projected simultaneously from the top of a `100 - m` high tower. Stone `B` is projected horizontally with speed `10 ms^-1`, and stone `A` is droppd from the tower. Find out the following :
(a) Time of flight of the two stone
(b) Distance between two stones after `3 s`
(c ) Angle of strike with ground
(d) Horizontal range of particle `B`.

Explanation

(a) To calculate the time of flight (for both stone), apply equation of motion in y - direction,
`100 = 1//2 "gt"^2`
`t = 2 sqrt(5) s`
(b) `x_B = 10 xx 3 = 30 mt`
`y_B = 1//2 xx g xx t^2`
=`1//2 xx 10 xx 3 xx 3`
=`45 m`
Distance between two stones sfter `3 s. x_B = 30 m`,
`y_B = 45 m`.
( c) Angle of striking with ground,
`v_y^2 = u_y^2 + 2 gh = 0 + 2 xx 10 xx 100`
`v_y = 20 sqrt(5) m s^-1 , v_x = 10 ms^-1`
`:. vec v = v_x hat i + v_y hat j`
`tan theta = (v_y)/(v_x)`
`rArr theta = tan^-1 ((20sqrt(5))/(10)) = tan^-1 (2 sqrt(5))`.
(d) Horizontal range of particle `B`,
`x_B = 10 xx (2 sqrt(5)) = 20 sqrt(5) m`.
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