1.

A ball thrown by one player reaches the other in 2 seconds. The maximum height attained by the ball above the point of projection will be (g = 10m//s^2)

Answer»

10m
7.5 m
5m
2.5m

Solution :Given 2 = `(2U sin theta)//G` or `(usin theta)//g =1`
max height `(u^2 sin^2 theta)/(2g)g//2` or `((u^2sin^2 theta)//g^2) 10/2xx1^2 = 5m`


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