Stumped on Physics Question?

Question

Physics Problem Trouble?
A seagull, ascending straight upward at 4.40 m/s, drops a shell when it is 15.5 m above the ground.

The magnitude and direction of the shell's acceleration just after it is released is 9.8 m/s^2 downward.

The maximum height above the ground reached by the shell is 16. 487755 m.

A. How long does it take for the shell to reach the ground?

B. What is the speed of the shell at this time?

HELP I am totally at a loss! I got that the max height is 16.487755, but I can't figure out the next two parts.

Nope, the two first answers are not right according to WebAssign. I am thinking that it has something to do with it goes from 15.5 m to 16.487755 m then falls to 0 m. Any ideas?

Answer 1

First off, you are correct with the maximum height being 16.487755 m (mind you, there is no need for the extra decimal places as you can only be as accurate as 3 significant figures).

If you use the Vf = Vi +at formula, you can the determine the time it takes to reach its maximum height as 0.449 seconds.

Then using the S=1/2at^2 +Vi*t formula you can determine the time it takes from its maximum height to reach the ground.

Let's work it out: 16.487755m = 1/2(9.8)t^2
(notice how our initial velocity is 0, that is because we are using our initial position as the shell's maximum height where it had no velocity)

we solve for t being as being 1.83 s, making our total time to be 2.28 s with significant figures.

for the second part of your problem, we can use our determined time to figure out our final velocity.

It is best to use the 'Vf=Vi +at' formula

note: for simplicity, we shall set our initial position at the maximum height so as to have an initial velocity of 0 m/s, this will mean that our time will be 1.83 s

so: Vf = 0 + 9.8(1.83)

Vf = 17.9 m/s in the downward direction

hope this helped

Answer 2

Since the shell is max height, it has an initial velocity of 0. Then to drop 16.48.... m,
x = 1/2 gt^2, where x is 16.48 and g is 9.8 m/sec2. This should be about 1.4 sec.

The speed when it hits the ground is v= gt,
where g is 9.8 m/sec^2 and t = 1.4 sec.

Answer 3

1. S = 0.5 g t^2. S and g are given; t can be solved for.
2. Use the time from the preceding part, and simply multiply by g.

Answer 4

in case you like to sq. a million then the respond is a million proper? yet once you like to respond to (a million/10)^ 2 then the glaring answer is a million^2/10^2 and the respond is a million/one hundred. a million/10 = 0.a million and a million/one hundred = 0.01. 0.9 is likewise 9/10 and the answer to sq. of 0.9 is 9^2/10^2 = 80 one/one hundred = 0.80 one that's why.