A 500g stone is thrown up with velocity of 5ms^-1. Find its
i) P.E at maximum hieght.
ii) K.E at its hit the ground.
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A 500g stone is thrown up with velocity of 5ms^-1. Find its
i) P.E at maximum hieght.
ii) K.E at its hit the ground.
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Verified answer
Given:-
→ Mass of the stone = 500g = 0.5kg
→ Initial velocity of the stone = 5m/s
To find:-
→ P.E. of the stone at maximum height.
→ K.E. of the stone as it hits the ground.
Solution:-
Number (a) :-
Firstly, let's calculate the maximum height reached by the stone by using the 3rd equation of motion :-
=> v² - u² = 2gh
=> 0 - (5)² = 2(-10)h
=> -25 = -20h
=> h = -25/-20
=> h = 1.25m
Now, we know that :-
P.E. = mgh
=> P.E. = 0.5(10)(1.25)
=> P.E. = 6.25 J
Number (b) :-
After reaching the maximum height, the stone will fall downwards [with u' = 0] So, let's calculate the velocity with which the stone hits the ground (v') by using the 3rd equation of motion :-
=> (v')² - (u')² = 2gh
=> (v')² - 0 = 2(10)(1.25)
=> (v')² = 25
=> v' = 5 m/s
Now, we know that :-
K.E. = 1/2mv²
=> K.E. = 1/2(0.5)(5)(5)
=> K.E. = 12.5/2
=> K.E. = 6.25 J
Thus :-
• P.E. at maximum height is 6.25J
• K.E. as it hits the ground is 6.25J .