Sum

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.

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#### Solution

Given:

Height of the cliff, h = 40 m

Initial speed of the projectile, u = 50 m/s

Let the projectile hit the ground with velocity 'v'.

Applying the law of conservation of energy,

\[\text{ mgh } + \frac{1}{2}\text{mu}^2 = \frac{1}{2}\text{mv}^2 \]

\[ \Rightarrow 10 \times 40 + \left( \frac{1}{2} \right) \times 2500 = \frac{1}{2} \text{v}^2 \]

\[ \Rightarrow \text{v}^2 = 3300\]

\[ \Rightarrow \text{ v = 57 . 4 m/s = 58 m/s} \]

The projectile hits the ground with a speed of 58 m/s.

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