In the figure shown, find
(a) the time when the particle strikes the ground at P.
(b) the horizontal distance QP
(c) velocity of the particle at P.

Answer

A, B, C, D

Explanation

(a)`t= sqrt((2h)/(g)) = sqrt ((2xx45)/(10)) = 3s`
(b) Horizontal distance `QP = 40t`
`= 40 xx 3 = 120m`
(c) Horizontal component of velocity at P = `40 m//s`
Vertical compound of velocity `= g t = 10t = 10 xx 3 = 30 m//s`

Net velocity v `= sqrt ((40)^2 + (30)^(2)) = 50 m//s`
`tan theta = 30/40 or 3/4 `
`:. theta = tan ^-1 (3/4) = 37^@` .