Physics, 25.09.2020 23:01 youngaidan21

5) [Honors]A seagull, ascending straight upward at 5.2 m/s, drops a shell when it is 12.5m above the ground. (A) What is the acceleration of the shell right after it is released? (B) Find the maximum height above the ground
reached by the shell. (C) How long does it take for the shell to return to a height of 12.5m above the ground?

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For the seagull that is ascending straight upward at 5.2 m/s and drops a shell when it is 12.5 m above the ground, we have:

A) The acceleration of the shell right after it is released is given by the acceleration due to gravity = 9.81 m/s².

B) The maximum height above the ground reached by the shell is 13.9 m.

C) The time for the shell to return to a heigh of 12.5 m above the ground is 1.06 s.

A) Since the shell is dropped, the acceleration after it is released is the same at each point of height as it falls, so the acceleration of the shell is:

$g = 9.81 m/s^{2}$

Where g is the acceleration due to gravity

Hence, the acceleration of the shell is 9.81 m/s²

B) The maximum height reached by the shell can be calculated with the following equation:

$h_{max} = h_{i} + h_{f}$ (1)

Where:

$h_{i}$ : is the initial height = 12.5 m

$h_{f}$ : is the final height

The final height is the height reached by the shell after it is released. It can be calculated with the next equation:

$v_{f}^{2} = v_{i}^{2} - 2gh_{f}$ (2)

Where:

$v_{f}$ : is the final velocity = 0 (in the maximum height)

$v_{i}$ : is the initial velocity = 5.2 m/s

$h_{f}$ : is the height reached by the shell after it is released

The final heigh is (eq 2):

$h_{f} = \frac{v_{i}^{2}}{2g} = \frac{(5.2 m/s)^{2}}{2*9.81 m/s^{2}} = 1.4 m$

Now, the maximum height is (eq 1):

$h_{max} = h_{i} + h_{f} = 12.5 m + 1.4 m = 13.9 m$

Then, the maximum height reached by the shell is 13.9 m.

C) The time for the shell to go from the maximum height to 12.5 m (falling time) can be calculated with:

$y_{f} = y_{i} + v_{i}t - \frac{1}{2}gt^{2}$

Where:

$v_{i}$ : is the initial velocity in its way down = 0 (it is going free fall)

$y_{f}$ : is the final height = 12.5 m

$y_{i}$ : is the initial height = 13.9 m (maximum height)

The falling time is:

$12.5 m = 13.9 m + 0 - \frac{1}{2}9.81 m/s^{2}*t_{f}^{2}$

$t_{f} = \sqrt{\frac{2*(13.9 m - 12.5 m)}{9.81 m/s^{2}}} = 0.53 s$

Now, the time to return to 12.5 is twice the above time (time to go up plus the time to go down).

$t = 2*t_{f} = 2*0.53 s = 1.06 s$

Therefore, the time for the shell to return to a height of 12.5 m is 1.06 s.

Find more about free fall here link

I hope it helps you!

Answer from: Jwelch4171

(B) 13.9 m

Explanation:

Given:

v₀ = 5.2 m/s

y₀ = 12.5 m

(A) The acceleration in free fall is -9.8 m/s².

(B) At maximum height, v = 0 m/s.

v² = v₀² + 2aΔy

(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)

y = 13.9 m

v = at + v₀

-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s

t = 1.06 s

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