Question

At what angle a ball should be thrown with a velocity of $24m{s}^1$ just to cross a wall $16m$ high at a horizontal distance of $32m$ . Given: $g=10m{s}^2$

Answer 1

using equation of trajectory for projectile motion

$y=xtan\theta -\frac{g{x}^2}{2u^{2}co{s}^2\theta }$

$y=xtan\theta -\frac{g{x}^{2}se{c}^2\theta }{2u^2}$

$y=xtan\theta -\frac{g{x}^2(1+ta{n}^2\theta )}{2u^2}$

using$y=16,x=32,u=24$ simplyfying above equation

$5ta{n}^2\theta -18tan\theta +14=0$

solving above equation

we get two solution $\theta ={48.6}^0$ and $\theta ={67.9}^o$