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# 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. - Physics

ConceptVarious Forms of Energy : the Law of Conservation of Energy

#### 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

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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Solution 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. Concept: Various Forms of Energy : the Law of Conservation of Energy.
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