1.

A projectile is fired at 45° with a speed of 200 ms^(-1) . Its maximum height will be the same as that for a projectile fired vertically upwards with a speed of:

Answer»

`400ms^(-1)`
`200sqrt(2)ms^(-1)`
`200/sqrt(2)ms^(-1)`
`100ms^(-1)`

Solution :Here `(u^(2)sin^(2)theta)/(2g)=(u^(2)sin^(2)45^@)/(2g)=u^(2)/(4g)`
Now for vertical THROW `h_(max)=v^(2)/(2g)`
As given `^(2)/(2g)=u^(2)/(4g)=(200xx200)/(4g)`
or `v^(2)=200xx100`
`v=200/sqrt(2)ms^(-1)`


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