Calculate the height from which a body is released from rest if its velocity just before hitting the ground is 30m/s.
Answers
Answer:
We can use conservation of energy principle to get the height. Let the height be h metre above the ground for it to attain the downward velocity of 30m/s just prior to hitting the ground.
The potential energy of the body at height h = m g h, where m is mass of the body and g is the acceleration due to gravity.
The kinectic energy of the body just before hitting the ground = ½ m v², where v is the velocity acquired by the body in free fall from height h. The value of v = 30 m/s
By conservation of energy principle, mgh = ½ m v².
Or
h = v²/2 g = 30 × 30/ 2 × 10 = 45 m, taking g = 10 m/s².
Verfication: When dropped from a height of 45m, the velocity attained by the body just before the ground can be obtained using
v² - u²= 2 × g× h; u= 0 m/s (as the body is just dropped). Therefore,
v² = 2 × 10 × 45 = 900 ; ==> v = 30 m/s.
So, when dropped from a height of 45 m and taking g = 10 m/s², the body would attain a velocity of 30 m/s just before hitting the ground.
Answer:
What is the height from which a body is released from rest of its velocity just before hitting the ground is 30m/s?
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What is the height from which a body be released from rest so that its velocity just before hitting the ground is 30 m/s?
We can use conservation of energy principle to get the height. Let the height be h metre above the ground for it to attain the downward velocity of 30m/s just prior to hitting the ground.
The potential energy of the body at height h = m g h, where m is mass of the body and g is the acceleration due to gravity.
The kinectic energy of the body just before hitting the ground = ½ m v², where v is the velocity acquired by the body in free fall from height h. The value of v = 30 m/s
By conservation of energy principle, mgh = ½ m v².
Or
h = v²/2 g = 30 × 30/ 2 × 10 = 45 m, taking g = 10 m/s².
Verfication: When dropped from a height of 45m, the velocity attained by the body just before the ground can be obtained using
v² - u²= 2 × g× h; u= 0 m/s (as the body is just dropped). Therefore,
v² = 2 × 10 × 45 = 900 ; ==> v = 30 m/s.
So, when dropped from a height of 45 m and taking g = 10 m/s², the body would attain a velocity of 30 m/s just before hitting the ground.
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Almost any projectile motion question can be solved using one of these three Newton’s equations of motion:
S=Vit+12at2 ————--eqn 1
Vf=Vi+at —————eqn 2
combine equation 1 and equation 2 to eliminate “t” gives
V2f−V2i=2aS —————eqn 3
It is highly recommended to watch your signs in these equations. Velocities are up = positive, down = negative and the acceleration due to gravity always points down so ay=−9.81m/s2
For your question, we are given:
Vi=0 and Vf=−30ms Note the negative sign which indicates a downward direction. You are asked to find S , so choose equation 3:
V2f−V2i=2aS
(−30)2−0=2(−9.81)S
S=−45.9m
Answer reported with only 3 significant figure accuracy because I used g=9.81 which has only 3 sig fig accuracy.
The negative sign means “below the starting point”, so the body will have a downward velocity of 30ms when dropped from a height of 45.9 m