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

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4 years ago

Vikas TU
14149 Points
```							500 = u^2*sin(2theta)/g................(1)and10 = u^2*(sin(theta))^2/2g..................(2)solving both eqns. we get,ucos(theta) = 50 andusin(theta) = 50we  get u = root(5000)Maximum height = u^2/2g = > 5000/20 = > 250 meter.
```
4 years ago
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