A body is projecte at an angle of `60^(@)` with the horizontal such that the vertical component of its initial velocity is `40 ms^(-1)` . The magnitude of velocity of the projectile at one quarter of its time of flight is nearly,
(Acceleration due to gravity, `g = 10 ms^(-2))`

Options

A. `3 . 54 ms^(-1)`
B. `35.40 ms^(-1)`
C. `30.54 ms^(-1)`
D. `34.5 ms^(-1)`

Answer

A

Explanation

According to the question,
Verticle component of velocity , `u_(y) = (sqrt(3))/(2) u = 40`
or `u = (80)/(sqrt(3))ms^(-1)`
Time of fight, T ` = (2 u sin theta)/(g)`
`= (2((80)/(sqrt(3)))sin 60^(@))/(10)`
`rArr T = 2 xx (80)/(sqrt(3)) xx (sqrt(3))/(2) xx (1)/(10) rArr T = 8 sec`
at time , `t = (T)/(4) = (8)/(4) = 2 sec`
`vx= ((80)/(sqrt(3)))/(2)= (40)/(sqrt(3)) ms^(-1)`
at `t = (T)/(4) = 2sec` ,
From first equation of the motion,
`v_(y) = u_(y) - gt = (sqrt(3))/(2)((80)/(sqrt(3))) - 10 xx 2 " "[:. u_(y) = " using " 60^(@)]`
`= 40 - 10 xx 2 = 20 ms^(-1)`
Hence, at t `=(T)/(4)` , magnitude of valocity of projectile
`v = sqrt(v_(x)^(2) +v_(y)^(2)) = ` Verticle component of velocity , `u_(y) = (sqrt(3))/(2) u = 40`
or `u = (80)/(sqrt(3))ms^(-1)`
Time of fight, T ` = (2 u sin theta)/(g)`
`= (2((80)/(sqrt(3)))sin 60^(@))/(10)`
`rArr T = 2 xx (80)/(sqrt(3)) xx (sqrt(3))/(2) xx (1)/(10) rArr T = 8 sec`
at time , `t = (T)/(4) = (8)/(4) = 2 sec`
`vx= ((80)/(sqrt(3)))/(2)= (40)/(sqrt(3)) ms^(-1)`
at `t = (T)/(4) = 2sec` ,
From first equation of the motion,
`v_(y) = u_(y) - gt = (sqrt(3))/(2)((80)/(sqrt(3))) - 10 xx 2 " "[:. u_(y) = " using " 60^(@)]`
`= 40 - 10 xx 2 = 20 ms^(-1)`
Hence, at t `=(T)/(4)` , magnitude of valocity of projectile
`v = sqrt(v_(x)^(2) +v_(y)^(2)) = sqrt(((40)/(sqrt(3)))^(2)+(20)^(2))~~30.54ms^(-1)`