a shell is fired with a velocity Vo at an angle theta with the horizontal derive the expressions for i) maximum height ii) horizontal range.
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Answer:
v^2=u^2+2as
Here, v=0 (final velocity), a=-g ( because body is moving against gravity) and s=H. u = u sin theta (along vertical component).
0= u^2 sin^2 theta - 2gH
H = u^2 sin^2 theta / 2g, hence proved.
Explanation:
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