Question

a ball is thrown with velocity 28 m/s and at an angle 30^ofrom the x axis. calculate the distance from the thrower to the point where the ball comes to the same level.

Answer 1

Given,
$$\begin{gathered} \mathrm{Velocity}\mathrm{of}\mathrm{projection},{\mathrm{v}}_0=28\mathrm{m}/\mathrm{s} \\ \mathrm{Angle}\mathrm{of}\mathrm{projection},{\mathrm{\theta }}_0={30}^0 \\ \mathrm{The}\mathrm{distance}\mathrm{from}\mathrm{the}\mathrm{thrower}\mathrm{to}\mathrm{the}\mathrm{point}\mathrm{where}\mathrm{the}\mathrm{ball}\mathrm{returns}\mathrm{to}\mathrm{the}\mathrm{same}\mathrm{level}\mathrm{is}\mathrm{given}\mathrm{by}, \\ \frac{{{\mathrm{v}}_0}^2\sin 2{\mathrm{\theta }}_0}{\mathrm{g}}=\frac{28\times 28\times \sin (2\times {30}^0)}{9.8} \\ =\frac{28\times 28\times \sin {60}^0}{9.8}=\frac{28\times 28\times \sqrt{3}}{2\times 9.8} \\ =69.3\mathrm{m} \end{gathered}$$