Before working this problem, review Conceptual Example 14. A pellet gun is fired straight downward from the edge of a cliff that is 14.7 m above the ground. The pellet strikes the ground with a speed of 27.2 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward

23.04 m

Explanation:

We'll begin by calculating the initial velocity of the pellet. This can be obtained as follow:

Height (h) of cliff = 14.7 m

Final velocity (v) = 27.2 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =?

v² = u² + 2gh

27.2² = u² + (2 × 9.8 × 14.7)

739.84 = u² + 288.12

Collect like terms

u² = 739.84 – 288.12

u² = 451.72

Take the square root of both side

u = √451.72

u = 21.25 m/s

Thus, the initial velocity of the pellet is 21.25 m/s.

Finally, we shall determine the maximum height to which the pellet would have gone assuming the gun was fired straight upward. This can be obtained as follow:

Initial velocity (u) = 21.25 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 m/s (at maximum height)

Maximum height (h) =?

v² = u² – 2gh (since the pellet is going against gravity.

0² = 21.25² – (2 × 9.8 × h)

0 = 451.5625 – 19.6h

Collect like terms

0 – 451.5625 = –19.6h

–451.5625 = –19.6h

Divide both side by –19.6

h = –451.5625 / –19.6

h = 23.04 m

Therefore, the pellet will reach a maximum height of 23.04 m above the cliff.