A ball thrown by one player reaches the other in 2 sec . The maximum height attained by the ball above the point of projection will be about
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Time of fall in oblique projection is
T = 2uSinθ / g
2 = 2uSinθ / g
uSinθ = g ………[1]
Maximum height attained by it is
H = u²Sin²θ/(2g)
= (uSinθ)² / (2g)
= g² / (2g) ………[∵ From Equation (1)]
= g/2
= 9.8 / 2
= 4.9 m
Maximum height attained by ball is 4.9 m
T = 2uSinθ / g
2 = 2uSinθ / g
uSinθ = g ………[1]
Maximum height attained by it is
H = u²Sin²θ/(2g)
= (uSinθ)² / (2g)
= g² / (2g) ………[∵ From Equation (1)]
= g/2
= 9.8 / 2
= 4.9 m
Maximum height attained by ball is 4.9 m
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