Solving a Vertical Throw Problem: Calculating Time, Speed, & Distance

[converse48]

Homework Statement

A stone is thrown vertically upward with a speed of 13.5 m/s from the edge of a cliff 74.0 m high
(a) How much later does it reach the bottom of the cliff?
(b) What is its speed just before hitting?
(c) What total distance did it travel?

Homework Equations

i am only able to use these two equations
Δx=1/2at^2+Vo
Vo=initial velocity(i guess you guys already know that)

and
Vf(final velocity)=at+Vo

The Attempt at a Solution

for c, first thing i did was get 1.38s by using the final velocity equation
and i put that and all the other data into the delta x equation
Δx=1/2(-9.8)(1.38^2)+13.5(1.38)
Δx=-4.9(1.90)+13.5(1.38)
Δx=9.32m
so i added that to 74 to get 83.32
BUT ITS WRONG!

 

[MysticDude]

Okay, I'm assuming that you are using g = 9.8m/s² for your acceleration.

I would divide this problem into two parts, one for the time that it is going upward, then the other for the part when the stone is falling downwards.

For (a). We would have to find the time that it takes the stone to have a Vf of 0 because at that point it will start to fall. In other words...t=\frac{v_f - v_0}{a} Since we know that for the first part t = -13.5/-9.8 = 1.38s (That is the time for it to go up and stop, then start falling down!) So JUST for the first part, it will take 2.76s for the stone to reach the start of the cliff again. Now we have to find out the second part and add the two up to get the total time. Now we have to use the second equation to get the rest of the time. We have to use the second equation because we don't know the v_{final} when the stone hits the ground. So -74 = \frac{1}{2}at^2, and some playing around with the formula we get \frac{2*-74}{-9.8} = t^2 so taking the square root of that (to get the time) we get t = 3.89s. So adding all of the time up, we get t_{total} = 6.65s
Note, I too am only in a class, so if there are please correct them :D.

 

[zgozvrm]

So -74 = \frac{1}{2}at^2, and some playing around with the formula we get \frac{2*-74}{-9.8} = t^2 so taking the square root of that (to get the time) we get t = 3.89s.

Close, but you're assuming that V_0 = 0[/tex] for the second part of the equation, which can't be: The stone reached it's peak and stopped at t \approx 1.38s[/tex], then <i>accelerated</i> downward from there, so it had to have <i>some</i> speed > 0 m/s when it once again reached the top of the cliff.

 

[MysticDude]

Close, but you're assuming that V_0 = 0[/tex] for the second part of the equation, which can't be: The stone reached it's peak and stopped at t \approx 1.38s[/tex], then <i>accelerated</i> downward from there, so it had to have <i>some</i> speed > 0 m/s when it once again reached the top of the cliff.

<br /> <br /> Oh okay thanks for clearing that up, really appreciate it :D. It seems as if this user is not going to reply, but thanks for telling me about my mistake...thanks again :D