A ball of 5 kg is thrown upwards with a speed of 10m/s
‹a› find the potential energy when it reach the highest point
‹B› calculate the maximum hight attained by it
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(B) Using the equation of motion:
Final Velocity ^2 = Initial Velocity^2 + 2* acceleration * Displacement
V^2 = U^2 + 2*a*s
Given, U = - 10 m/s
Mass = 5kg
Since the velocity will be zero at the highest point, the final velocity (V) =0
0 = 10^2 + 2(-10) * displacement
0 = 100 (-20* displacement)
-100 = - 20 * displacement
- 100 ÷ - 20 = displacement
Therefore, Displacement = 5 meters
So, the maximum height attained by it is 5 meters.
(A) Potential Energy = mgh
= mass * g * height
= 5 * 10 * 5
= 250 Joules
Therefore, the Potential Energy at the highest point is 250 Joules.
Final Velocity ^2 = Initial Velocity^2 + 2* acceleration * Displacement
V^2 = U^2 + 2*a*s
Given, U = - 10 m/s
Mass = 5kg
Since the velocity will be zero at the highest point, the final velocity (V) =0
0 = 10^2 + 2(-10) * displacement
0 = 100 (-20* displacement)
-100 = - 20 * displacement
- 100 ÷ - 20 = displacement
Therefore, Displacement = 5 meters
So, the maximum height attained by it is 5 meters.
(A) Potential Energy = mgh
= mass * g * height
= 5 * 10 * 5
= 250 Joules
Therefore, the Potential Energy at the highest point is 250 Joules.
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