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Answer:
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.