[111] Throw-up problem - Launch/freefall

[WPCareyDevil]

Homework Statement

A stone is launched straight up by a slingshot. Its initial speed is 19.8 m/s and the stone is 1.40 m above the ground when launched. Assume g = 9.80 m/s2.
(a) How high above the ground does the stone rise?
wrong check mark m
(b) How much time elapses before the stone hits the ground?

Homework Equations

D= Vot + (1/2)at^2
V=at+Vo

The Attempt at a Solution

Vi=19.8m/s
d=1.4m
a=-9.8m/s

I actually solved another question I had while typing it up on here.. maybe this one will be the same.

Ok, I've attacked this thing every way I can think of, and my problem is the integration of it being launched from 1.4m above the ground. I tried 1.4=19.8t+ (1/2)(-9.8)t^2, solving quadratically for t=3.9688 and t=.071987, neither of which are correct. I know finding the time is the key to the solution, but I am not sure how to integrate the 1.4m into the formula.

Can someone point me in the right direction?

 

[BuGzlToOnl]

Let me rewrite this your formula for part A is y= (v^2-v0^2)/(2a).

After you get your y component your formula for Par B is: t= v-v0/a.

t = time
a = acceleration
y = distance (vertical)

And that should give you the answer for the problem.

Sorry went through 3 or so edits kept copying the wrong equation, that should be it thou. Times the answer by 2 to get the up time and downtime for Part B.

 

[baba89]

I would first make sure that all the given information is accurate and consistent. I would double check the initial velocity and height, as well as the acceleration due to gravity. I would also check the units to ensure they are consistent.

Next, I would approach the problem by breaking it down into smaller parts. First, I would find the time it takes for the stone to reach its maximum height. This can be done by using the equation d=Vot + (1/2)at^2, where d is the displacement (in this case, the maximum height), Vo is the initial velocity, and a is the acceleration due to gravity. Solving for t will give us the time it takes for the stone to reach its maximum height.

Next, I would find the maximum height by plugging in the time obtained in the previous step into the equation d=Vot + (1/2)at^2. This will give us the displacement at the maximum height.

Finally, I would use the same equation to find the time it takes for the stone to reach the ground. This time, however, the initial displacement would be the maximum height obtained in the previous step. This will give us the total time it takes for the stone to travel from its initial position to the ground.

In this way, we can break down the problem into smaller, manageable parts and use the given equations to solve for the required values.