Two boys conducted experiments on the projectile motion with stopwatch and noted some readings. As one body throws a stone in air at the same angle with the horizontal, the other boy observes that after 4 s, then stone is moving at an angle `30^(@)` to the horizontal and after another 2 s it is travelling horizontally. the magnitude of the initial velocity of the stone is (Acceleration due to gravity,
g = 10 `ms^(-2)`. )

Options

A. `40 sqrt(3) ms^(-1)`
B. `20sqrt(3) ms^(-1)`
C. `10sqrt(3) ms^(-1)`
D. `50sqrt(3) ms^(-1)`

Answer

A

Explanation

Given acceleration due to gravity, g = 10 `m//s^(2)`
After 4s, angle between stone and horizontal plane, `theta = 30^(@)`
After t = 4s, equation of the vertical projectile motion, when `theta = 30^(@)`
After t = 4s, equation of the vertical projectile motion, when `theta = 30^(@)`
`therefore " " tan theta = (v sin theta - g(f))/(v cos theta) " "`.....(i)
`tan 30^(@) = (vsin theta - g(4))/(v cos theta)" "`.....(ii)
total time to reach the stone at horizontal surface, t = 2 + 4 = 6 s
After t = 6 s, equation of horizontal projectile motion, `theta = 0^(@)`,
`tan 0^(@) = (v sin theta - g(6))/(v xx cos 0 ) " " [ ( therefore cos 0^(@) = 1),(tan 0^(@) = 0)]`
or `" " v sin theta` -g(6) = 0
v sin `theta = 60 " "`.....(iii)
`" "` [ `therefore` Given, g = 10]
from Eq. (ii), At t = 4 s, when particles travelling in horizontal direaction,
` therefore " " v cos theta = 20sqrt(3) " "` ....(iv)
Now, magnitude of initial velocity,
` v = sqrt(( v sin theta)^(2) + (V cos theta)^(2))`
`" "` [ From Eqs. (iii) and (iv) ]
or `" " v = sqrt((60)^(2) + (20 sqrt(3))^(2))`
or `" " v = 40 sqrt(3)` m/s
So, the magnitude of initial velocity of stone is
v = `40sqrt(3)` m/s .