Question

a ball of mass 2kg is thrown upward vertically with a speed of 4m/s find the potential energy when it reaches. a) find the potential energy when it reaches the highest point, b) calculate the maximum height it reaches​

Answer 1

Answer:

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Answer 2

Answer:

Let the ball goes to h meter height.

Now mass of the ball =m = 2 kg

The speed in ground is = v= 4 m/s

So kinetic energy(KE) is

$KE = \frac{1}{2} m {v}^{2}$

At maximum height the all kinetic energy converted to potential energy. So we get

$mgh \: = \frac{1}{2} m {v}^{2} \\ or \: . \: h = \frac{ {v}^{2} }{2g}$

$so. \: h = \frac{ {4}^{2} }{2 \times 9.8} \\ \: \: \: \: \: = 0.82 \: m$

The potential energy at maximum height is the kinetic energy in the ground. So the potential energy(PE) at max. Point is

$PE = mgh \: = 2 \times 9.8 \times 0.82 \\ \: \: \: \: = 16 \: joule$