1.

For an object thrown at `45^(@)` to the horizontal, the maximum height H and horizontal range R are related asA. `R = 16H`B. `R = 8H`C. `R = 4H`D. `R = 2H`

Answer» Correct Answer - C
As we know that maximum height of a projectile is given by
`H_(max) = (u^(2)sin^(2)theta)/(2g)`
where, u = initial velocity of projectile
g = acceleration due to gravity
and `theta=` angle of projection.
As from question,
`H_(max) = (v^(2)sin^(2)45^(@))/(2g)`.....(i) (as `u =v, theta = 45^(@))`
Now, range of a projectile is given by
`R = (u^(2)sin 2 theta)/(g) rArr R = (v^(2)sin (2xx 45^(@)))/(g)`
`rArr R = (v^(@) sin 90^(@))/(g)` .......(ii)
On dividing Eq. (i) by Eq. (ii). we get
`(H_(max))/(R) = (v^(2)sin^(2)45^(@) xx g)/(2g xx v^(2)sin 90^(@)) = (1)/(4 xx1)`
`rArr R = 4H_(max) = 4H`


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