Question

Phil throws a ball upwards that weighs 55 Newtons (N). The ball travels at a speed of 4 meters per second (m/s) and reaches its maximum height at 9.5 meters (m). At what height is the potential energy of the ball equal to the kinetic energy of the ball? Round to the nearest hundredth if necessary.

Answer 1

Given:

Weight of ball (m) = 55N

Speed of ball (u) = 4m/s

Max height (h) reached by ball= 9.5m

To find:

Height where KE = PE

Solution:

Let x metres be the height at which KE = PE

since total energy is constant

(KE + PE) at height x = (KE + PE) at max height h

or

$\frac{1}{2} m {v}^{2} + \: mgx \: = 0 \: + mgh$

(At max height, velocity is 0m/s so KE at h = 0)

(v is velocity at height x)

Also, given that u = 4m/s

and taking g= 9.8m/s^2

using equation of motion

${v}^{2} = {u}^{2} + 2as$

or v^2 = 16+19.6x

Substituting,

$\frac{1}{2} \times 55 \times (16 + 19.6x) + 55 \times 9.8 \times x = 55 \times 9.8 \times 9.5$

solving for x,

x= 4.34m

Hence, at 4.34m Potential energy of ball is equal to Kinetic energy of ball.

Answer 2

Given : Phil throws a ball upwards that weighs 55 Newtons (N).

The ball travels at a speed of 4 meters per second (m/s) and reaches its maximum height at 9.5 meters (m).

To Find : At what height is the potential energy of the ball equal to the kinetic energy of the ball?

Solution:

TE = KE + PE

TE = Total Energy

KE = Kinetic Energy = (1/2) mv²

PE = Potential Energy = mgh

Phil throws a ball upwards that weighs 55 Newtons (N).

reaches its maximum height at 9.5 meters (m).

At max height v = 0  

Hence KE = 0    

PE = mgh

mg = 55N (weight)

h = 9.5

PE = 55 * 9.5 = 522.5  Joule

TE = PE + KE  = 0 + 522.5  = 522.5 Joule

When PE = KE

=> PE = KE = TE/2

=> PE = 522.5/2

522.5/2 = 55 * h

=> h  = 9.5/2

=> h = 4.75 m

At Height 4.75 m  the potential energy of the ball equal to the kinetic energy of the ball

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