1. ## projectile

a projectile is fired with velocity u at an angle a above the horizontal.. it lands on the same horizontal plane at point P. it lands the furthest point away on the plane. show that for the projectile to hit a point H above point P and its projected at the same angle the initial velocity must be u^2/root(u^2-gh)

i know that the angle of elevation is 45 degrees. im trying to use ut-.5gt^2=h

2. ## Re: projectile

1) Use $u$ to determine the range of the first projectile

2) write the equation of motion for the second projectile that travels with initial speed $w$

3) a point on the curve given by (2) is (range, H), thus you have two equations, one in each of the components

4) solve these two equations for $w$ and $t$

5) note that you want the answer where $t>0$

3. ## Re: projectile

For the initial projectile landing at point $P$ ...

$\Delta x = \dfrac{v_0^2 \sin(2\theta)}{g} = \dfrac{u^2}{g}$

The second projectile launched at the higher speed (call it $v$) will have the same value for $\Delta x$ when it reaches the height, $H$, directly above point $P$.

$\Delta x= v\cos(45) \cdot t = \dfrac{u^2}{g} \implies t = \dfrac{u^2\sqrt{2}}{gv}$

substitute the above expression for $t$ into the equation $H=v\sin(45) \cdot t - \dfrac{1}{2}gt^2$, then solve for $v$.

4. ## Re: projectile

thanks i get it now. i worked the distance in the x direction in a different way though