A seagull, ascending straight upward at 6.25 m/s, drops a shell when it is 17.5 m above the ground.?

Question

a) What is the magnitude and direction of the shell's acceleration just 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 reach the ground?
d) What is the speed of the shell at this time?

Answer 1

A) acceleration due to gravity on earth is always 9.81m/s^2 DOWN so that's the answer unless the seagull isnt on earth.



B) The shell is dropped when the seagull is 17.5m above the ground. Since gravity acts downwards, it cant go up once it is released. So the answer is 17.5... unless you have to like subtract 17.5 from the height of the seagull or something.



C) The initial velocity of the shell is 0, since the only force is acting on it after it's released. Acceleration is -9.81 (negative means down) and the distance is 17.5. You have 3 knowns and an unknown now you can find an equation on the back of the ref table and do your math.



D) The speed it hits the ground with is the final velocity. Acceleration is change in velocity (vf-vi) over time. A = (vf-vi) / t Plug in the numbers and there you have it.



Have fun doing your math ;P

Answer 2

As the seagull moves 17.5 meters upward, its acceleration is -9.8 m/s^2. This means the seagull’s velocity will decrease from 6.25 m/s to 0 m/s as it rises to its maximum height. Use the following equation to determine the distance the seagull rises.



vf^2 = vi^2 + 2 * a * d, a = -9.8 m/s^2

0 = 6.25^2 + 2 * -9.8 * d

-6.25^2 = -19.6 * d

d = 6.25^2 ÷ 19.6 ≈ 2 meter

This is the distance the seagull moves upward.



Maximum height = 17.5 + 2 = 19.5

The maximum height is approximately 19.5 meters. As the egg falls this distance, it velocity increases from 0 m/s to the final velocity at the rate of 9.8 m/s each second. Use the following equation to determine the time.



d = vi * t + ½ * 9.8 * t^2, vi = 0

19.5 = 4.9 * t^2

t = √(19.5 ÷ 4.9)

The time is approximately 2 seconds



Final velocity = 0 + 9.8 * 2 = 19.6 m/s

This is the final speed.

Answer 3

v(t) = at + v0

h(t) = (1/2)at^2 + v0 * t + h0



a) 9.8 m/s^2 downward (gravity)



b) At the maximum height, v(t) = 0,

v(t) = 0 = -9.8 * t + 6.25 Solve for t

h(t) = (1/2) * (-9.8) * t^2 + 6.25 * t + 17.5 Evaluate at the value of t from above.



c) h(t) = 0 = (1/2) * (-9.8) * t^2 + 6.25 * t + 17.5 Solve for t



d) v(t) = -9.8t + 6.25 Evaluate for t found in part c.