Question

A pellet gun is fired straight downward from the edge of a cliff that is 12 m above the ground. The pellet strikes the ground with a speed of 26 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?

Answer 1

Please help. I am confused.

Answer 2

First you need is the initial velocity. So we put the kinematic equations,
\[v=v_0-gt\\
y=y_0+v_0t-\frac{1}{2}gt^2\]In this case,
\[v_0=v+gt\\
y=y_0+(v+gt)t-\frac{1}{2}gt^2=y_0+vt+\frac{1}{2}gt^2\]Using the data,
\[0=12-26t-4.9t^2\Rightarrow t=0.427\ s\]And,
\[v_0=v+gt=-26+9.8\cdot0.427=-21.82\ m/s\]
For the second part, first we calculate the time the bullet will fly,
\[v=0=v_0-gt\Rightarrow t=\frac{v_0}{g}=\frac{21.82}{9.8}=2.23\ s\]
As you see, now the initial velocity is positive because is fired in the positive senses of the y vertical direction. Calculate the height the bullet will reach above the cliff,
\[h-h_0=v_0t-\frac{1}{2}gt^2=24.3 \ m\]