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A stone of mass 0.1kg is thrown vertically upwards at a speed of 5.0m/s from a 20m high...
A stone of mass 0.1kg is thrown vertically upwards at a speed of 5.0m/s from a 20m high building.
Ignoring the effect of air resistance,
1. Calculate the potential energy of the stone before it is thrown
2. What is the maximum height the stone reaches above the hand of the thrower?
1 Answer | add yours
Lets assume the ground level as the datum.
Initial potential energy
= 19.62 J
At the moment of the release of the stone it has the potential energy due to its height from datum and the kinematic energy due its velocity. When it reaches the maximum height the kinematic energy will vanish and potential energy will increase.
If we assume the height of the stone reached above the building as H;
Using law of energy conservation
Final potential energy = Initial potential and kinematic energy
`0.1*g*(20+H) = 1/2*0.1*5^2+0.1*g*20`
` H = 1.274m`
So the stone will reach 1.274m above the hand of the thrower.
Posted by jeew-m on September 24, 2012 at 4:13 PM (Answer #1)
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