# Find the maximum horizontal range of a cricket ball projected with a velocity of 80m/s. If the ball is to have a range of 1003 meters, find the least angle of projection and the least time taken.

Unicorn
96 Points
2 years ago
For maximum range angle of projection = 45
Range max=u^2/g=80^2/10=640 meters

If range =1003
1003=80^2 sin2 theta/g
Sin2 theta =1003/640
Theta=1/2 sin^-1(1003/640)

Time =2 u sintheta /g
::))
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29 Points
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