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Oksanka [162]
3 years ago
13

A canon is tilled back 30.0 degrees and shoots a cannon ball at 155 m/s. What is the

Physics
1 answer:
Umnica [9.8K]3 years ago
6 0

Answer:

= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m

Therefore, highest point that the cannon ball reaches is 168.7m

Explanation:

the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s

highest point that the cannon ball reaches?

H_{max}=\frac{V^2\sin ^2 \theta}{2g}

g = 9.8m/s2

= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m

Therefore, highest point that the cannon ball reaches is 168.7m

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