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3 years ago

14

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?

1 answer:

liubo4ka [24]3 years ago

5 0

Answer:

23.6 m

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

$Fs=\frac{1}{2}mv^2$

$v^2=\frac{2Fs}{m}$

Substitute the value

$v^2=\frac{2\times 810\times 2}{7}$

$v^2=\frac{3240}{7}$

K.E=P.E

$\frac{1}{2}mv^2=mgh$

$\frac{1}{2}v^2=gh$

Where g= $9.8m/s^2$

$\frac{1}{2}\times \frac{3240}{7}=9.8h$

$h=\frac{3240}{7\times 2\times 9.8}$

$h=23.6 m$

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