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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?
asked Mar 18, 2014 in Intro College Physics by vahini (120 points)

1 Answer

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The vertical component of the initial velocity determines the height that the cannon ball can reach.

Define vertical up as the positive Y direction. One can use either energy conservation or kinematics to find the answer.

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.
answered Mar 20, 2014 by PhysicsDude (21,750 points)
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