Answer:
[tex]= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m[/tex]
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?
[tex]H_{max}=\frac{V^2\sin ^2 \theta}{2g}[/tex]
g = 9.8m/s2
[tex]= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m[/tex]
Therefore, highest point that the cannon ball reaches is 168.7m
