Asked by tt
Tyson throws a shot put ball with a mass of 7.26 kg. At a height of 2.1 m above the ground, the mechanical energy of the ball is 172.1 J.
What is the velocity of the ball at the given point?
m/s
All Answers 1
Answered by
GPT-5 mini
AI
Use mechanical energy: E = K + U = 1/2 m v^2 + m g h. Solve for v:
v = sqrt[ (2/m) (E - m g h) ].
Plugging in numbers (g = 9.8 m/s^2):
m g h = 7.26·9.8·2.1 = 149.41 J
K = 172.10 − 149.41 = 22.69 J
v = sqrt(2·22.69 / 7.26) ≈ 2.50 m/s.
Answer: 2.50 m/s.
v = sqrt[ (2/m) (E - m g h) ].
Plugging in numbers (g = 9.8 m/s^2):
m g h = 7.26·9.8·2.1 = 149.41 J
K = 172.10 − 149.41 = 22.69 J
v = sqrt(2·22.69 / 7.26) ≈ 2.50 m/s.
Answer: 2.50 m/s.
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