Question

During a contest that involved throwing a 7.0-kg bowling ball straight up in the air, one contestant exerted a force of 810 N on the ball. If the force was exerted through a distance of 2.0 m, how high did the ball go from the point of release?

Answer 1

To find the height the bowling ball reached, we can use the work-energy principle. The ball went approximately 23.2 meters high from the point of release.

Step-by-step explanation:

To determine the height the bowling ball reached, we can use the work-energy principle. The work done on the ball is equal to the change in its potential energy. The work done can be calculated using the formula:

Work = Force x Distance

Substituting the given values:

Work = 810 N x 2.0 m = 1620 J

The potential energy of the ball at its highest point is equal to the work done on it:

Potential Energy = Work = 1620 J

Using the formula for gravitational potential energy:

Potential Energy = mgh

where m is the mass of the ball (7.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height we want to find. Rearranging the formula:

h = Potential Energy / (mg)

Substituting the given values:

h = 1620 J / (7.0 kg x 9.8 m/s^2) ≈ 23.2 m

Therefore, the ball went approximately 23.2 meters high from the point of release.

Answer 2

23.6 m

Step-by-step explanation:

We are given that

Mass of ball, m=7.0 kg

Force, F=810 N

Distance, s=2.0 m

We have to find the height of ball from the point where it releases.

Work done=K.E

`Fs=(1)/(2)mv^2`

`v^2=(2Fs)/(m)`

Substitute the value

`v^2=(2* 810* 2)/(7)`

`v^2=(3240)/(7)`

K.E=P.E

`(1)/(2)mv^2=mgh`

`(1)/(2)v^2=gh`

Where g=
`9.8m/s^2`

`(1)/(2)* (3240)/(7)=9.8h`

`h=(3240)/(7* 2* 9.8)`

`h=23.6 m`