A particle is projected from ground with velocity `40 m//s` at `60^@` from horizontal.
(a) Find the speed when velocity of the particle makes an angle of `37^@` from horizontal.
(b) Find the time for the above situation.
(C ) Find the vertical height and horizontal distance of the particle from the starting point in the above position. Take `g= 10m//s^2` .

Answer

A

Explanation

In the figure shown,
`u_x = 40 cos 60^@ = 20 m//s `
`u_y = 40sin60^@ = 20sqrt(3) m//s`
`a_x =0 and a_y = -10m//s^2`
(a) Horizontal component of velocity remains unchanged.

`v_x = u_x`
or `vcos37^@ = 20 `
`rArr v(0.8)=20`
`:. v = 25 m//s`.
(b) Using, `v_y = u_y + a_yt rArr t=(v_y-u_y)/(a_y)`
`For t_1, t_1 = (vsin37^@-20sqrt(3))/(-10) = ((25)(0.6)-20sqrt(3))/(-10) = 1.96s`
` For t_2, t_2 (-vsin37^@-20sqrt(3))/(-10) = (-(25)(0.6)-20sqrt(3))/(-10) = 4.96s`
(c) Vertical height `h=s_y`
Let us calculate at `t_1`
`:. h=u_y t_1 +1/2 a_y t_(1)^(2) = (20sqrt(3))(1.96) + 1/2 (-10)(1.96)^2`
=48.7 m.
Horizontal distance
`x_1 = u_x t_1`
= (20)(1.96) = 39.2m
Similarly, `x_2 = u_xt_2 = (20)(4.96)`
=99.2 m.