Solve the following :


A projectile is fired from the top of a 40 m high cliff with an initial speed of 50 m/s at an unknown angle. Find its speed when it hits the ground.


By law of conservation of energy

$\mathrm{mgh}+\frac{1}{2} \mathrm{~m} \mathrm{v}_{\mathrm{i}}^{2}=\frac{1}{2} \mathrm{mv}_{\mathrm{f}}$

$10 \times 40+\frac{1}{2} \times 50^{2}=\frac{1}{2} \times V_{f}^{2}$

$v=v_{f}=57.4 \mathrm{~m} / \mathrm{s}$

$v \approx 58 \mathrm{~m} / \mathrm{s}$


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