A body is projected at such an angle that the horizontal range is three times the greatest height. If the range is 400metres, find:
A. the necessary velocity of the
projection
B. the time of flight. ( take g=10m/s)
PLEASE I NEED THE ANSWER AS FAST AS POSSIBLE
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its greatest height: 400/3m
consider vertical motion: consider horizontal motion:
v^2=u^2+2gs v=s/t
u^2=2*10*400/3 v=400/5.16=77.5m/s
u=51.6m/s v=root(77.5^2+51.6^2)=93.1m/s
v=u+gt
0=u-10t
t=5.16s
so time=5.16s, velocity=93.1m/s
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