A body is thrown with velocity (4i+3j)m/s.Its maximum height
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Answered by
22
the initial velocity is (4i+3j).
so its initial velocity in y direction is 3m/s.
at maximum height, its final velocity is zero.
so using v^2=u^2-2gh
0=3^2 - 2(10)h
20h=9
h=9/20m=0.45m
so its initial velocity in y direction is 3m/s.
at maximum height, its final velocity is zero.
so using v^2=u^2-2gh
0=3^2 - 2(10)h
20h=9
h=9/20m=0.45m
Answered by
2
Explanation:
0.45
by using the formula u^2sin^2theta/2g
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