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: $H_{MAX}=\frac{{v_0}^2{\sin }^2\theta }{2g}$ 
or,         $H_1=\frac{{v_0}^{2}si{n}^2@}{2g} eq 1$ 


and  $$\begin{gathered} H_2=\frac{{v_0}^{2}si{n}^2(90-@)}{2g} \\ or H_2=\frac{{v_0}^2{\cos }^2@}{2g} eq. 2 \end{gathered}$$

By adding eq 1 & 2 we get

 $\frac{{v_0}^2}{2g}=H_1+H_2$    
or , ${v_0}^2=2g(H_1+H_2)$

Now, Total Time of Projectile motion

vertical displacement y=0      and vertical component of velocity= v₀sin@ 
or, $0=v_0\sin @*T -\frac{1g{T}^2}{2}$
  $T=\frac{2v_0\sin @}{g}$         eq. 3

Now horizontal range is given by using horizontal component of velocity i.e. v_0​cos@

or R=v_0​cos@*T

    from eq.3 we have

$$\begin{gathered} R=\frac{v_0\cos @ x 2 v_0 \sin @}{g} \\ or R = \frac{{v_0}^2\sin (2@)}{g} as \sin 2@= 2\sin @\cos @ \end{gathered}$$   

This is same for case angle @ measured from virtical.


    putting in value of v₀2

$$\begin{gathered} R = \frac{2g(H_1+H_2)sin(2@)}{g} \\ or R = 2(H_1+H_2)*sin(2@) \end{gathered}$$


Regards.