A ball of mass 4 kg is thrown vertically upwards from the ground with the initial velocity of 40 m/sec. Find the change in potential energy of the ball when it reaches maximum height. (Neglect air friction)
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mass = 5 kg
u = -10 m/s
v = 0 m/s
g = -10 m/s²…for this case
2as = v² – u²
2(-10)(s) = -100
S = -100/-20
S = 5 m
Now,
Potential energy = mgh
P.E = 5 * 10 * 5
P.E = 250 J
Now, Kinetic energy
K.E = 1/2mv²
K.E = ½*5*100
K.E = 5*50
K.E = 250 J
I hope it helps
u = -10 m/s
v = 0 m/s
g = -10 m/s²…for this case
2as = v² – u²
2(-10)(s) = -100
S = -100/-20
S = 5 m
Now,
Potential energy = mgh
P.E = 5 * 10 * 5
P.E = 250 J
Now, Kinetic energy
K.E = 1/2mv²
K.E = ½*5*100
K.E = 5*50
K.E = 250 J
I hope it helps
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