Question:
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.
Solution:
By law of conservation of energy
$\mathrm{mgh}+\frac{1}{2} \mathrm{~m} \mathrm{v}_{\mathrm{i}}^{2}=\frac{1}{2} \mathrm{mv}_{\mathrm{f}}$
$10 \times 40+\frac{1}{2} \times 50^{2}=\frac{1}{2} \times V_{f}^{2}$
$v=v_{f}=57.4 \mathrm{~m} / \mathrm{s}$
$v \approx 58 \mathrm{~m} / \mathrm{s}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.