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LovePhys

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## Homework Statement

A cannon ball is fired with a velocity of 30m/s at an angle of 35° from the top of a cliff. The height of the cliff is 50m. Calculate total time of flight of the cannon ball before it hits the ground.

[itex] v_o=30m/s [/itex]

[itex] \alpha=35° [/itex]

[itex] g=10m/s^2 [/itex]

## Homework Equations

The equation of the parabola expressing the orbit of the cannon ball:

[itex] y= \frac {-gx^2}{2v_o^2cos^2\alpha} + tan(\alpha) x [/itex]

## The Attempt at a Solution

Obviously, a part of the parabola is missing as the cannon ball is struck from the cliff. I chose the position of the cannon as the origin and put a coordinate system there.

I have 2 ways of solving this problem:

1/ Calculate the time taken for the cannon ball to reach the maximum height:

[itex] t=\frac {v_osin(\alpha)}{g} ≈ 1.7 (s) [/itex]

The maximum height is: [tex] h=\frac{v_o^2 \sin^2(\alpha)}{2g}≈15(m) [/tex]

Then, pretending that the cannon ball is struck vertically from the height of 15+50=65(m), the time taken for it to hit the ground is:

[tex] \sqrt{\frac{2.65}{g}}≈3.6(s) [/tex]

So that the total time is about: 1.7+3.6=5.3(s)

2/ I want to place the cannon at a position that allows me to see the full parabola. I worked out the quadratic equation of the parabola, and I also had the new horizontal range of the cannon ball if the cannon is at a new position. However, here is where it get stuck. I do not know the time taken for the cannon ball to get from the new position to the original position, also the new angle at which it was thrown, and the new velocity.

Can anyone please help me with this?

I really appreciate your help!

Thanks in advance!

LovePhys