# a) calculate the maximum height above the ground the canon ball reaches?

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a canon ball is fired from an artillery dugout such that initial height of the canon ball is 0.0m (ground level). the canon ball has an initial velocity of 200m/s at 55 angle to the horizontal as shown below. you can neglect air resistance.

a) calculate the maximum height  above the ground the canon ball reaches?

Energy conservation: $$E_{initial}=E_{final}$$ $$KE_i+PE_i=KE_f+PE_f$$ $$\frac12mV_i^2+0=\frac12mV_x^2+mgh$$
​Note here $V_x=V_icos(55^\circ)$ is a constant and h is the maximum height it can reach. Then we have $$\frac12m(V_i^2-V_x^2)=mgh$$
The Kinematics way: $$V_{fy}^2-V_{iy}^2=2gh$$ which gives $$0-V_{iy}^2=2(-9.8)h$$
Here $V_{iy}=V_isin(55^\circ)$ and $g=-9.8m/s^2$. Note the direction of the g is in the downward direction and therefore is negative.