Question

A ball is thrown at angle $\theta$ and another ball is thrown at angle $({90}^{\circ }-\theta )$ with horizontal direction from the same point each with speeds of $40m/s$. The second ball reaches $50m$ higher than the first ball. Find their individual heights, $g=10m/s^2$.

Answer 1

Let ball $B_1$ is thrown at angle ${\theta }_1$ & ball $B_2$ is thrown at $\angle {\theta }_2=\angle (90-{\theta }_1)$

Initial velocity, $V=40m/s$

Height achieved by ball $B_1=H_1$

& height achieved by ball $B_2=H_2$

Now, $H_1=\frac{V^2{\sin }^2{\theta }_1}{2g}=\frac{{\left(40\right)}^2{\sin }^2{\theta }_1}{2\times 10}⟶\left(i\right)$

$H_2=\frac{V^2{\sin }^2(90-{\theta }_1)}{2g}$

$=\frac{V^2{\cos }^2{\theta }_1}{2g}=\frac{{\left(40\right)}^2\times {\cos }^2{\theta }_1}{2\times 10}⟶\left(ii\right)$

As $H_2=H_1+50m$

$\Rightarrow {80\cos }^2{\theta }_1=80{\sin }^2{\theta }_1+50$

$\Rightarrow (1-{\cos }^2{\theta }_1)=80{\sin }^2{\theta }_1+50$

$\Rightarrow 8-8{\sin }^2{\theta }_1=8{\sin }^2{\theta }_1+5$

$\Rightarrow 3=16{\sin }^2{\theta }_1⟶\left(iii\right)$

& ${\cos }^2{\theta }_1=\frac{13}{16}⟶\left(iv\right)$

Putting $\left(iii\right)$ in $\left(i\right)$ & $\left(iv\right)$ in $\left(ii\right)$

$\Rightarrow H_1=\frac{40\times 40\times 3}{20\times 16}$, $H_2=\frac{40\times 40\times 13}{20\times 16}$

$H_1=15m$ $H_1=65m$

Height achieved by ball $B_1=15m$

& height achieved by ball $B_2=65m$.