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

Options

A. `400ms^(-1)`
B. `200sqrt(2)ms^(-1)`
C. `200/sqrt(2)ms^(-1)`
D. `100ms^(-1)`

Answer

C

Explanation

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)`