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.
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

HEY!!

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✴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 + 1/2 mu^2 = 1/2mv^2

▶10 × 40 + (1/2)× 2500= 1/2v^2

▶v^2= 3300

▶v = 57.4 m/s = 58 m/s

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

Answer 2

Given in the question , 
Height h = 40 m.
Initial velocity u = 50 m/s.

Now potential of projectile w.r.t the ground at the time it projected,
P.E= mgh
P.E= mg × 40 J

K.E when projected = 1/2 mv²
= m × 1250 J.

Now total energy at time of the projection ,
Kinetic Energy + Potential Energy
(mg × 40) + ( m × 1250) J.

assume that the speed of the projectile when it comes to ground = V
Then Kinetic energy = 1/2 mV² J
So Total energy = K.E + P.E = 1/2 mV² 
1/2 mV² = (mg × 40 ) + (m × 1250 ) J
V² = 80 × g + 2500
V² = 80 × 9.8 + 2500
V²= 3284
V =√3284
V= 57.30 m/s


Hence the speed when it hit ground is 57.30 m/s.


Hope it Helps.