1.

A stone is thrown vertically at a speed of 30 ms^(-1) taking an angle of 45^(@) with the horizontal. What is the maximum height reached by the stone ? Take g = 10 ms^(-2).

Answer»

30 m
22.5 m
15 m
10 m

Solution :`H_("max") = (v^(2) SIN^(2) THETA)/(2g)`
`H_("max") = (30^(2) sin^(2)(45))/(2 xx 10)`
`H_("max") = 22.5 m`


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