Question

A boy tossed a ball of which mass is 50 g straight up with an initial velocity 50
m/s. Find the potential and kinetic Energy at 4s after the ball tossed.

Answer 1

Answer:

s it is  at maximum height it will have only potential energy no kinetic energy because of zero velocity. Hence P.E=mgh

But when it is on the ground it has no height from the ground hence P.E is zero where K.E has a value of mgh because energy is conserved at every point.

And substitute values given to you

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Answer 2

Given :

• Mass of ball (m) = 50 g = 0.05 kg
• Initial velocity (u) = 50 m/s
• Time interval (t) = 4 seconds
• Acceleration due to gravity (a) = -10 m/s²

To Find :

• Potential Energy
• Kinetic Energy

Solution :

First of all we have to find the height up to which ball is reached. So, use 2nd equation of motion,

$\implies \sf{s \: = \: ut \: + \: \dfrac{1}{2} at^2} \\ \\ \implies \sf{s \: = \: 50 \: \times \: 4 \: + \: \dfrac{1}{2} \: \times \: -10 \: \times \: 4^2} \\ \\ \implies \sf{s \: = \: 200 \: + \: -5 \: \times \: 16} \\ \\ \implies \sf{s \: = \: 200 \: - \: 80} \\ \\ \implies \sf{s \: = \: 120 \: m}$

$\therefore$ Height up-to which ball is reached is 120 m

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Now, we will find out Potential Energy :

$\implies \sf{P.E \: = \: mgh} \\ \\ \implies \sf{P.E \: = \: 50 \: \times \: 10 \: \times \: 120} \\ \\ \implies \sf{P.E \: = \: 60000 \: J}$

$\therefore$ Potential Energy will be 60000 Joules

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Use 1st equation of motion :

$\implies \sf{v \: = \: u \: + \: at} \\ \\ \implies \sf{v \: = \: 50 \: + \: 4 \: \times \: -10} \\ \\ \implies \sf{v \: = \: 50 \: - \: 40} \\ \\ \implies \sf{v \: = \: 10}$

$\therefore$ Final velocity is 10 m/s

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Now, we have to find Kinetic Energy :

$\implies \sf{K.E \: = \: \dfrac{1}{2} mv^2} \\ \\ \implies \sf{K.E \: = \: \dfrac{1}{2} \: \times \: 0.05 \: \times \: 10^2} \\ \\ \implies \sf{K.E \: = \: \dfrac{1}{2} \: \times \: 0.05 \: \times \: 100} \\ \\ \implies \sf{K.E \: = \: 2.5}$

$\therefore$ Kinetic Energy of Ball is 2.5 Joules