A projectile is fired horizontally with velocity of 98 m/s from the top of a hill 490 m high. Find
(a) the time taken by the projectile to reach the ground,
(b) the distance of the point where the particle hits the ground from foot of the hill and
(c) the velocity with which the projectile hits the ground. `(g= 9.8 m//s^2)`
.

Explanation

In this problem we cannot apply the formula of R,H and T directly. Here, it will be more convenient to choose x and y directions as shown in figure.
Here, `u_(x) = 98 ms^(-1), a_(x) = 0, u_(y) = 0` and `a_(y) = g`
(i) At `A, s_(y) = 490 m`. So, applying
`s_(y) = u_(y)t +(1)/(2)a_(Y)t^(2)`
`:. 400 = 0 +(1)/(2) (9.8) t^(2) :. t = 10 s`
`BA = s_(x) = u_(x)t +(1)/(2)a_(x)t^(2)` or `BA = (98) (10)+0 ( :' a_(x) = 0)`
or `BA = 980 m`
(iii) Horizontal velocity `v_(x) = u_(x) = 98 ms^(-1)`
Vertical velocity `v_(y) = u_(y) +a_(y)t`
`= 0 +(9.8) (10) = 98 ms^(-1)`
`:.` Resultant velocity `v = sqrt(v_(x)^(2)+v_(y)^(2)) = sqrt((98)^(2) +(98)^(2))`
`= 98 sqrt(2) ms^(-1)`
and `tan beta = (v_(y))/(v_(x)) = (98)/(98) = 1 :. beta = 45^(@)`
Thus, the projectile hits the ground with a velocity `98 sqrt(2) ms^(-1)` at an angle of `beta = 45^(@)` with horizontal as shown in the above figure.