Question

A cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 70.0m above ground level, and the ball is fired with initial horizontal speed v0. Assume acceleration due to gravity to be g = 9.80m/s2 .

Part A

Assume that the cannon is fired at time t=0 and that the cannonball hits the ground at time tg. What is the y position of the cannonball at the time tg/2?

Answer is 52.5

Part B

Given that the projectile lands at a distance D = 200m from the cliff, as shown in the figure, find the initial speed of the projectile, v0.

Express the initial speed numerically in meters per second.

v0 =

52.9

m/s

Part C

What is the y position of the cannonball when it is at distance D/2 from the hill? If you need to, you can use the trajectory equation for this projectile, which gives y in terms of x directly:

y=H?gx22v20x.

You should already know v0x from the previous part.

Express the position of the cannonball numerically in meters.

yD/2 =		?????? m

Answer 1

Use the below kinematic equation to find the \(y\) -position of the cannon ball at the time \(t_{g} / 2\)

$$ y-y_{0}=v_{0 y}-\frac{1}{2} g t_{g}^{2} $$

In this case, the initial vertical position of the cannonball is \(y_{0}=H\) and the final vertical position of the cannonball is \(y=0\) because it hits the ground finally. The initial vertical velocity of the cannonball is \(v_{0 y} .\) Thus, the above kinematic equation becomes as,

$$ \begin{aligned} 0-H &=0-\frac{1}{2} g t_{g}^{2} \\ H &=\frac{1}{2} g t_{g}^{2} \end{aligned} $$

Here, \(t_{g}\) is the time taken to reach the ground by the cannonball and \(g\) is the acceleration due to gravity.

Rewrite the equation for \(t_{g}\)

$$ t_{g}=\sqrt{\frac{2 H}{g}} $$

Therefore, the time taken to reach the ground by the cannonball is,

$$ \begin{aligned} t_{g} &=\sqrt{\frac{2(70.0 \mathrm{~m})}{9.8 \mathrm{~m} / \mathrm{s}^{2}}} \\ &=3.78 \mathrm{~s} \end{aligned} $$

Now, determine the y-position of the cannonball at the time \(t_{g} / 2\).

$$ \begin{aligned} y-y_{0} &=v_{0 y}-\frac{1}{2} g t_{g}^{2} \\ y-H &=0-\frac{1}{2} g\left(\frac{t_{g}}{2}\right)^{2} \\ y-H &=-\frac{1}{2} g\left(\frac{t_{g}}{2}\right)^{2} \\ y &=H-\frac{1}{2} g\left(\frac{t_{g}}{2}\right)^{2} \end{aligned} $$

Therefore, the \(y\) -position of the cannonball at the time \(t_{g} / 2\) is,

$$ \begin{aligned} y &=(70.0 \mathrm{~m})-\frac{1}{2}\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\left(\frac{3.78 \mathrm{~s}}{2}\right)^{2} \\ &=\mathbf{5 2 . 5} \mathrm{m} \end{aligned} $$