A ball is thrown horizontally with speed `20 m//s` from a tower of height `80 m`.
(a) After how much time and at what horizontal distance from the foot of tower it strikes the ground.
(b) Find the magnitude and direction of the velocity with which it strikes the ground.
( c) Find the velocity vector of ball after `1 s`.
(d) Trajectory followed by the ball.

Explanation

(a) `O` to `B`:
In the verticle direction:
`y = h = (1)/(2) g t^(2)`
`80 = (1)/(2) xx 10 t^(2) rArr t = 4 s`
In the horizontal dorection:
`x = R = ut = 20 xx 4 = 80 m`
The particle strikes the ground after `4 s` and at the horizontal distance `80 m` from the foot of tower.
(b) `O` to `B`: `v_(x) = u = 20 m//s`
`v_(y) = g t = 10 xx 4 = 40 m//s`
`v = sqrt(v_(x)^(2) + v_(y)^(2)) = sqrt((20)^(2) + (40)^(2)) = 20sqrt(5) m//s`
`tan alpha = (v_(y))/(v_(x)) = (40)/(20) rArr alpha = tan^(-1)(2)`
The particle strikes the gorund with speed `20sqrt(5) m//s` at angle `tan^(-1)(2)` with horizontal.
( c) After `1 s`, `v'_(x) = 20 m//s, v'_(y) = g t = 10 xx 1 = 10 m//s`
`vec(v) = v'_(x) hat(i) + v'_(y) hat(j) = 20 hat(i) + 10 hat(j) m//s`
(d) After time `t, x = ut = 20 t, y = (1)/(2) g t^(2) = 5 t^(2)`
Eliminating `t`, we get
`y = 5((x)/(20))^(2) = (x^(2))/(80)` , `y = (x^(2))/(80) ("parabola")`