Answered

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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?

Sagot :

lublana

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

[tex]Fs=\frac{1}{2}mv^2[/tex]

[tex]v^2=\frac{2Fs}{m}[/tex]

Substitute the value

[tex]v^2=\frac{2\times 810\times 2}{7}[/tex]

[tex]v^2=\frac{3240}{7}[/tex]

K.E=P.E

[tex]\frac{1}{2}mv^2=mgh[/tex]

[tex]\frac{1}{2}v^2=gh[/tex]

Where g=[tex]9.8m/s^2[/tex]

[tex]\frac{1}{2}\times \frac{3240}{7}=9.8h[/tex]

[tex]h=\frac{3240}{7\times 2\times 9.8}[/tex]

[tex]h=23.6 m[/tex]

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