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Grade: 12th pass

                        

Range of projectile motion =500mtime=10secondfind maximum height?

4 years ago

Answers : (1)

Vikas TU
12281 Points
							
500 = u^2*sin(2theta)/g................(1)
and
10 = u^2*(sin(theta))^2/2g..................(2)
solving both eqns. we get,
ucos(theta) = 50 
and
usin(theta) = 50
we  get u = root(5000)
Maximum height = u^2/2g = > 5000/20 = > 250 meter.
3 years ago
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