# A stone projected from ground with certain speed at a angle @ With horizontal attains maximum height H1,when it is projected with same speed at an angle @ with vertical attains height H2.The horizontal range of projectile is .?

Dear Student,

Max. height of a projectile: ${H}_{MAX}=\frac{{{v}_{0}}^{2}{\mathrm{sin}}^{2}\theta }{2g}$
or,

and

By adding eq 1 & 2 we get

$\frac{{{v}_{0}}^{2}}{2g}={H}_{1}+{H}_{2}$
or , ${{v}_{0}}^{2}=2g\left({H}_{1}+{H}_{2}\right)$

Now, Total Time of Projectile motion

vertical displacement y=0      and vertical component of velocity= v0sin@
or,
$T=\frac{2{v}_{0}\mathrm{sin}@}{g}$         eq. 3

Now horizontal range is given by using horizontal component of velocity i.e. v0cos@

or R=v0cos@*T

from eq.3 we have

This is same for case angle @ measured from virtical.

putting in value of v02

Regards.

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