A person is standing on a truck moving with a constant velocity of 14.7 m/s o a hrozontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection. a. as seen from the truck b. as seen fromt the road.

Answer

A, B, C, D

Explanation

As seen from the truck the ball moves vertically upward comes back.
Time taken time taken by truck to cover 58.8 m.
`:. Time = S/v = 58.8/14.7 = 4 sec`

Hencer `v= 14.7 m/s for truck`
`u=? , v=0= -9.8 m/s^2 `
(going upward)
`t= 4/2 = 2sec`
`:. v= u+at`
`rarr 0=u+9.8x2`
`rarr = 19.6 m/s`
(vertical upward velocity)
b. From road, it seems to be projectivel motion.
tota time of flight = 4 sec
Horizontal range covered in this time
`= 58.8 m=X`
`:. x= u cos thetas t`
rarr u cos theta t = 14.7`
Taking vertical component of veocity into consideration.
`y= (0^2-(19.6)^2)/(2xx(-9.8))`
`= 19.6m ` {from (a)}
`:. y= u sin theta t- 1/2 gt^2`
`rarr 19.6 = u sin theta (2)- 1/2 (9.8) 2^2`
`rarr 2u sin theta=19.6xx2`
`rarr 2u sin theta = 19.6xx 2`
`rarr u sin theta = 19.6`
Dividing equation (2) by equatiion 1 we get
`(u sin theta)/(u cos theta) = tan theta = 19.6/14.7 = 1.333`
`rarr theta tan^-1 (1.333)=53^0`
Again, ` u cos theta = 14.7`
`rarr u = 14.7/(cos 53^0)= 24.42 m/s`
The speed of ball is 24.42 m/s at an angle `53^0` with horizontal as seen from the road.