Question

A body is thrown vertically upward at the speed of 10m/s . Calculate the maximum height using Principal of conservation of energy .

NOTE: Solve by using Principal of conservation of energy not by kinematics .

Class 11

Topic : Work energy and power.

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Answer 1

Hey Supriya Here is your answer.

Total Energy = KE +PE
=
$\frac{1}{2} m{v}^{2} + mgh$
At time t=0 (at ground)
h=0, v=10
At the maximum height
h=hmax v=0

applying conversation of total energy
we get
$\frac{1}{2} m {10}^{2} + 0 = 0 + mg(hmax) \\ 50 \div g = hmax \\ hmax = 5$
therefore maximum height= 5meter

hope it helps..
please marks me brainliest if you found it useful.

Answer 2

The moment you throw it, it will possess kinetic energy and zero potential energy as per your reference frame, thus solving According to your reference position we get

1/2*m*10*10 = m*10*x

We get x as 5 meters using kinematics will get the same conclusion as the third equation is the same as conservation of energy under constant acceleration.