Solving Kinematics homework

[Falinox]

Homework Statement

http://i754.photobucket.com/albums/xx183/reddiesel08/Phyque.jpg


2. The attempt at a solution
∆x = ∆x
v₀ + .5(a)(t²) = v(t)
.5(9.8)t² = 35t
4.9t² - 35t = 0 Using the quadratic formula I get
t = 7.95 sec

∆x = v(t)
35(7.95) = 278.25m

Need help setting up and solving the problem, thanks in advance to anyone willing to help :)

 

[SammyS]

35/4.9 ≈ 7.143 s.

Why use the quadratic formula? -- although it should give the right answer. (35 ± √(35²-0) )/(2(4.9))

 

[SteamKing]

You've calculated the distance the ball travels from the drop point to the point where it passes superman. The problem however asks you determine how far above the sidewalk the ball is when it passes superman.

 

[Falinox]

Thanks everyone! I was able to figure it out.

 

[rwisz]

I would first like to commend you for attempting to solve this kinematics problem on your own. It shows that you are actively engaging with the material and trying to understand it. Now, let's go through your solution and make sure we have everything set up correctly.

Firstly, it is important to clearly define all the variables in the problem. In this case, we have ∆x representing the displacement (or change in position), v₀ representing the initial velocity, a representing the acceleration, t representing time, and v(t) representing the final velocity.

Next, we need to identify the given values in the problem. From the given image, we can see that ∆x = 35m, v₀ = 0 (since the object is initially at rest), and a = 9.8m/s² (assuming we are on Earth).

Now, we can use the kinematic equations to set up our problem. The first equation, ∆x = v₀t + .5at², relates displacement, initial velocity, acceleration, and time. Plugging in our given values, we get:

35 = (0)t + .5(9.8)t²

This simplifies to:

4.9t² = 35

Next, we can use the second kinematic equation, v = v₀ + at, to find the final velocity. Plugging in our given values, we get:

v = 0 + (9.8)t

This simplifies to:

v(t) = 9.8t

Now, we can use our known value for ∆x and our equation for v(t) to solve for t. Plugging in our given value for ∆x, we get:

35 = 9.8t

Solving for t, we get:

t = 35/9.8 = 3.57 seconds

However, this value only gives us the time it takes for the object to reach the displacement of 35m. The problem asks for the total time it takes for the object to travel 35m and then return to its original position. This means we need to double our calculated time of 3.57 seconds, giving us a total time of 7.14 seconds.

Finally, we can use our calculated value for t to find the final velocity. Plugging in our calculated value for t into our equation for v(t), we get:

v