A projectile is fired at an upward angle of from the top of a 135-m-high cliff with a speed of. What will be its speed when it strikes the ground below? (Use conservation of energy.)

The total mechanical energy of a system remains conserved. If the total work done by all external and internal forces is zero, the mechanical energy remains constant.

It is given by:

\(P{E_{\rm{i}}} + K{E_{\rm{i}}} = P{E_{\rm{f}}} + K{E_{\rm{f}}}\)

Here,\(P{E_{\rm{i}}}\)is the initial potential energy,\(K{E_{\rm{i}}}\)is the initial kinetic energy,\(P{E_{\rm{f}}}\)is the final potential energy, and\(K{E_{\rm{f}}}\)is the final kinetic energy of the system.

The angle of projection of the projectile is\(\theta = {38^{\rm{o}}}\).

The height of the cliff is\({h_{\rm{i}}} = 135\;{\rm{m}}\).

The initial speed of the projectile is \({v_{\rm{i}}} = 165\;{\rm{m/s}}\).

The final height attained by the projectile after reaching the ground will be\({h_{\rm{f}}} = 0\). Therefore, the final potential energy will also be zero. By applying the law of conservation of mechanical energy at the top and bottom of the cliff, you get:

\(\begin{aligned}P{E_{\rm{i}}} + K{E_{\rm{i}}} &= 0 + K{E_{\rm{f}}}\\P{E_{\rm{i}}} + K{E_{\rm{i}}} &= K{E_{\rm{f}}}\\mg{h_{\rm{i}}} + \frac{1}{2}mv_{\rm{i}}^2 &= \frac{1}{2}mv_{\rm{f}}^2\end{aligned}\) … (i)

This implies that the kinetic energy at the bottom is equal to the sum of kinetic and potential energies at the top.

Rearrange equation (i) and substitute the known numerical values in it.

\(\begin{aligned}{v_{\rm{f}}} &= \sqrt {v_{\rm{i}}^2 + 2g{h_{\rm{i}}}} \\ &= \sqrt {{{\left( {165\;{\rm{m/s}}} \right)}^2} + 2\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {135\;{\rm{m}}} \right)} \\ &= 172.83\;{\rm{m/s}}\\ &\approx 173\;{\rm{m/s}}\end{aligned}\)

Thus, the speed of the projectile when it reaches the ground is \(173\;{\rm{m/s}}\).