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)`
.

Options

A. `12 s` , `980m` , `98(sqrt2) m//s`
`beta = 45`
B. `10 s` , `98m` , `98(sqrt2) m//s`
`beta = 45`
C. `10 s` , `980m` , `98 m//s`
`beta = 45`
D. `10 s` , `980m` , `98(sqrt2) m//s`
`beta = 45`

Answer

D

Explanation

Here, it will be more convenient to choose x and y
directions as shown in figure.
Here, `u_x = 98 m/s , a_x = 0, u_y =0 `and` a_y = g`
(a) At A, `s_y = 490 m`. So, applying
`s_y = u_(y)t + 1/2(a_y)t^2`
`:. 490 = 0 + 1/2(9.8)t^2 `
`:. t=10s `
(b) ` BA=s_x = u_xt + 1/2(a_x)t^2`
or ` BA = (98)(10)+0`
or `BA = 980m`
(c) `v_x = u_x + a_xt = 98+0 = 98 m//s`
`v_y = u_y + a_y t = 0+(9.8)(10)=98m//s`
`:. v = (sqrt((v_(x)^2)+(v_(y)^2))) = (sqrt((98)^2 + (98)^2)) = 98 (sqrt2) m//s`
and `tan beta = ((v_y)/(v_x)) = 98/98 = 1`
`:. beta = 45^@`
Thus, the projectile hits the ground with velocity `98(sqrt2) m//s` at an angle of `beta = 45^@` with horizontal as shown in Fig. 7.9.