Question : A ball is thrown by one player reaches the other in 2s. What is the maximum height attained by the ball above the point of projection?
Doubt by Avani
Solution :
Time of flight = 2 s
2usinθ/g = 2
2usinθ = 2g
usinθ = 2g/2
usingθ = g — (1)
2usinθ/g = 2
2usinθ = 2g
usinθ = 2g/2
usingθ = g — (1)
Maximum Height = ?
Hmax = u2sin2θ/2g
Hmax = (usinθ)2/2g
Hmax = g2/2g [From equation (1)]
Hmax = (usinθ)2/2g
Hmax = g2/2g [From equation (1)]
Hmax = g/2
Hmax = 4.9 m
Hmax = 4.9 m
Hence, the maximum height attained by the ball above the point of projection is equals to 4.9 m