Question

While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 6.57 m/s. The stone subsequently falls to the ground, which is 17.9 m below the point where the stone leaves your hand. At what speed does the stone impact the ground? How much time is the stone in the air? Ignore air resistance and take g = 9.81 m/s2. (This is NOT a suggestion to carry out such an experiment!)

Answer 1

@BAdhi

Answer 2

@whpalmer4

Answer 3

@Shalante

Answer 4

@kropot72

Answer 5

@iambatman

Answer 6

@whpalmer4

Answer 7

I need help finding the elapsed time.

Answer 8

@matt101

Answer 9

While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 6.57 m/s. The stone subsequently falls to the ground, which is 17.9 m below the point where the stone leaves your hand. At what speed does the stone impact the ground? How much time is the stone in the air? Ignore air resistance and take g = 9.81 m/s2. (This is NOT a suggestion to carry out such an experiment!)

Okay, you can write an equation for the height of the stone as a function of time:

\[h(t) = -\frac{1}{2}gt^2 +v_0 t + h_0\]where \(g=9.81 \text{ m/s}^2\), \(v_0 = 6.57 \text{ m/s}\) and \(h_0 = 17.9\text{ m}\)

as you can see, at \(t=0\) that gives you a height of \(17.9\) meters, with the ground at a height of \(0\) meters. To find the elapsed time, solve the equation for \(h(t) = 0\) and find the (positive) value of \(t\) that makes it true. I suggest solving symbolically and then plugging in numbers, rather than doing algebra with quantities like \(6.57t\). Once you know the time at which the stone hits the ground, you can find the speed at impact by taking the derivative of the height function with respect to time and evaluating it at that time. Have you had calculus? If not, you can also find it by finding the apex of the flight (hint: the vertex of the parabola), and then using the elapsed time from that point to impact to calculate speed, knowing that the acceleration is constant, so \(v = gt\) where \(t\) is the time spent falling from the peak of the flight.

Answer 10

the function I wrote has 3 terms, the first representing the motion due to gravity (hence the negative sign), the second representing the motion due to the initial velocity (positive), and the third representing the starting position.

Answer 11

I was able to figure out the impact speed but I'm having issues with finding the time.

Answer 12

I'm not sure what exactly I'm suppose to do to find the time.

Answer 13

how did you find the impact speed?

Answer 14

(6.57 m/s) +2(-9.81 m/s^2)(17.9)

Answer 15

*(6.57 m/s)^2

Answer 16

and that equals?

Answer 17

20.0 because you also take the square root.

Answer 18

uh, square root of a negative quantity?

Answer 19

That was a typo. *9.81

Answer 20

Sorry.

Answer 21

okay, I agree that you get the correct answer if you do that, but it isn't 20.0, even if rounded...

Answer 22

Would it be the 19.86 then?

Answer 23

anyhow, formula for height as a function of time is

\[h(t) = -\frac{1}{2}gt^2 + v_0 t + h_0\]as I mentioned, with \(g=9.81,\ v_0=6.57,\ h_0=19.7\) (units omitted)

Time of impact is the value of \(t\) which makes \(h(t)=0\).

Yes, 19.86 is better than 20.0.

Answer 24

You can use the quadratic formula to solve the equation:

\[a x^2 + b x + c = 0\]solutions are
\[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

You have \(a = -\frac{1}{2}g, \ b = v_0, \ c = h_0\)

Answer 25

sorry, 17.9, not 19.7!

Answer 26

=-6.57+/-\[\sqrt{6.57^2-4(-4.905)(17.9}\]

Answer 27

all of that divided by -4.905

Answer 28

right

Answer 29

13.29

Answer 30

13.29/-4.905 = -2.71

Answer 31

actually, forgot the \(2\) in \(2a\) for the denominator, so you divide by \(-9.81\)

Answer 32

you want the positive solution, so you need to take the negative value of the numerator

Answer 33

1.35

Answer 34

no...

\[\frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-6.57\pm\sqrt{(6.57)^2-4(-\frac{1}{2})(9.81)(17.9)}}{2*(-\frac{1}{2})(9.81)}\]\[=\frac{-6.57\pm\sqrt{(6.57)^2+2(9.81)(17.9)}}{-9.81}\]\[=\]

Answer 35

Here's a graph of height vs. time:

Answer 36

Sorry. I keep getting the same answer for some reason.

Answer 37

Well, it's 2.69404 or thereabouts, as you can see on the graph.

Answer 38

Looks like you are dividing by an extra factor of 2 or something.

Answer 39

When I add the -6.57 I get 1.35 and get -13.29 when I subtract it.

Answer 40

Expand the polynomial before you try to solve it.

\[0 = -\frac{1}{2}gt^2 +v_0 t + h_0\]\[0 = -(0.5)(9.81)t^2 + 6.57 t + 17.9\]\[0 = -4.905t^2 + 6.57t + 17.9\]\[a=-4.905, b =6.57, c = 17.9\]\[t = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-6.57\pm\sqrt{(-6.57)^2 -4(-4.905)(17.9)}}{2(-4.905)}\]\[=\frac{-6.57-\sqrt{43.1649+351.198}}{-9.81}\]

Answer 41

I got it this time. I was just entering it into my calculator wrong. Sorry about that.

Answer 42

Thanks for helping. Now on to the next problems.

Answer 43

Okay, good luck! I have to go now.

Answer 44

Alright. Thanks again