4. A 0.145 kg ball leaves a person\'s hand with an upward speed of 12.0 m/s. A)

Question

4. A 0.145 kg ball leaves a person's hand with an upward speed of 12.0 m/s. A) Draw a sketch of the situation. B) What is the initial kinetic energy of the ball? C) What is the initial potential energy of the ball? D) What is the final kinetic energy of the ball (at the top)? E) Use the conservation of energy to solve for the maximum height of the ball. 5. A 0.150 kg ball leaves a person's hand at rest straight down from 2.5 meters above the ground. A) Draw a sketch of the situation. B) What is the initial kinetic energy of the ball? C) What is the initial potential energy of the ball? D) Use the conservation of energy to solve for the speed of the ball just as it hits the ground.

Explanation / Answer

4. (B)Initial KE=(KE)i= 0.5*m*ui2 = 0.5*(0.145 kg)*(144 m/s) =10.44 J

(C) Initial PE= (PE)i= mgh = (0.145kg)*(9.8 m/s2)*(0 m) = 0 J (Since h = 0 m on ground)

(D) Final KE=(KE)f= 0.5*m*uf2 = 0.5*(0.145 kg)*(0 m/s) =0 J ( Since, on top, velocity = 0)

(E) By law of conservation of energy,

Change in KE = Change in PE

=> 0.5*m*ui2- 0.5*m*uf2 = (PE)f-(PE)i

=> 10.44 J -0 = mgh- 0 J; Where, h = maximum height reached by the body

=> h = 10.44 J/(m*g)

=> h= 10.44 J/(0.145kg*9.8 m/s2)

=> h= 7.34 m

5. (B)Initial KE=(KE)i= 0.5*m*ui2 = 0.5*(0.150 kg)*(0 m/s) =0 J (Since, the body is freely falling)

(C) Initial PE= (PE)i= mgh = (0.150kg)*(9.8 m/s2)*(2.5 m) = 3.675 J ( h = 2.5 m from ground)

(D) By law of conservation of energy,

Change in KE = Change in PE

=>0.5*m*ui2- 0.5*0.150 kg*uf2 = (PE)f - (PE)i

=> 0 J -7.50*uf2 = 0 J-3.675 J; Where, uf = final velocity at ground

=> uf = (3.675 /7.50)0.5 = 7 m/s

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