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Students were having a catapult contest, launching a tennis ball with their home built devices. The winner launched a ball at 30 degrees from the ground at 6.0 m/s. How high does the ball go?

Sagot :

lublana

Answer:

The height reached by the ball=0.459 m

Explanation:

We are given that

[tex]\theta=30^{\circ}[/tex]

Initial speed, u=6 m/s

We have to find the maximum height reached by ball.

Maximum height reached by ball

[tex]h=\frac{u^2sin^2\theta}{2g}[/tex]

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

Substitute the values

[tex]h=\frac{6^2sin^230^{\circ}}{2\times 9.8}[/tex]

[tex]h=\frac{36\times \frac{1}{4}}{2\times 9.8}[/tex]

[tex]h=0.459 m[/tex]

Hence, the height reached by the ball=0.459 m

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