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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.\t

Arnav Daryani , 3 Years ago

Grade 11

2 Answers

Unicorn

Last Activity: 3 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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Last Activity: 2 Years ago

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