Two graphs of the same projectile motion (in the xy-plane) projected from origin are shown. X-axis is along horizontal direction and y-axis is vertically upwards. Take `g = 10 ms^(-2)`.

Projection angle with the horizontal is :

Options

A. Y compenent of initial velocity is 5 m/s
B. X component of initial velocity is 4 m/s
C. Time to reach maximum height is 0.5 sec
D. All the above are correct

Answer

D

Explanation

From graph (A) : `V_(y) =0` at `t = (1)/(2) sec`
i.e., time taken to reach maximum height H is
`t = (u_(y))/(g) = (1)/(2) rArr u_(y) = 5 m//s`
from graph (B) : `v_(y) = 0` at `x =2 m`
i.e., when the particle is at maximum height, its displacement along horizontal `x = 2 m`
`x = u_(x) xx t rArr 2 = u_(x) xx (1)/(2)`
`rArr u_(x) = 4 m//s`