A ball is projected from the ground at a speed of `10ms^(-1)` making an angle of `30^(@)` with the horizontal. Another ball is simultaneously released from a point on the vertical line along the maximum height of the projectile.Both balls collide at maximum height. The initial height of the second ball is `(g=10ms^(-2))`

Options

A. 6.25 m
B. 2.5 m
C. 3.75 m
D. 5 m

Answer

B

Explanation

Maximum height of projectile
`h=(u^(2)sin^(2)theta)/(2g)`
`therefore h=((10)^(2)xxsin^(2)(30^(@)))/(2xx10)`
`=(5)/(4)=1.25m`
Time to reach maximum height
`t=(usintheta)/(g)`
`therefore t=(10xxsin30^(@))/(10)=0.5s`
So, distance of vertical fall in 0.5 s
`s=(1)/(2)g t^(2)`
or `s=(1)/(2)xx10xx(0.5)^(2)`
`=1.25m`
`therefore` Height of second ball
`=1.25+1.25=2.5m`