Question

The coach throws a baseball to a player with an initial speed of `20 m//s` at an angle of `45^@` with the horizontal. At the moment the ball is thrown, the player is 50 m from the coach. At what speed and in what direction must the player run to catch the ball at the same height at which it was released? `(g = 10 m//s^2)`.

Answer 1

Correct Answer - A::B
`T = (2usin theta)/g = ((2 xx 20 xx 1/sqrt2)/10) = 2(sqrt2)s`
`R = (u^2sin 2 theta)/g = ((20)^2sin90^@)/10`
Now the remaining horizontal distance is
`(50-40)m = 10m.` Let v is the speed of player,then
`vT = 10 or v=10/T = 10/(2sqrt2) = 5/(sqrt2) m//s`.