Question

a projectile is fired from the top of a 40 m height with an initial speed of 50 M per second at an unknown angle find its speed when it hits the ground

Answer 1

Hallo dear ♡

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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,

mgh+12mu2=12mv2⇒10 × 40 + (12)× 2500=12v2⇒v2=3300⇒ v=57.4 m/s=58 m/s

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

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Answer 2

Explanation:

given,

height of a body (h)= 40meter

initial velocity (u)= 50m/s

here acceleration = gravitational force because the body is falling downwards on ground so gravitational force acts here,

we know that ,gravitational force = 10 metre per second

we know that,

v²-u²=2as

v²=u²+2as

by substituting the values given

v²=(50)²+2(10)(40)

v²=2500+800

v²=3300

sending the ( )² to right side

v= √3300

v=√ (33)(100)

v= ( √33) (√100)

v=(5.8)(10)

v=58

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