Falling object 1D Kinematic Question

[CandyApples]

Homework Statement

An object falls from height h from rest. If it travels .5h in the last second before hitting the ground, determine how long it takes to fall and the height h.
xi = .5hm, t =1s, a = 9.8m/s^2, xf = 0m,

Homework Equations

Vf = Vi + at
xf = xi + .5at^2

The Attempt at a Solution

xf = xi + .5at^2
0 = .5h + .5(-9.8)(1)
h = 9.8m

I know that this answer is not correct, I am having trouble figuring out how to manipulate the given data into one of those equations to either come up with h or find the total time.

 

[Jebus_Chris]

Homework Statement

An object falls from height h from rest. If it travels .5h in the last second before hitting the ground, determine how long it takes to fall and the height h.
xi = .5hm, t =1s, a = 9.8m/s^2, xf = 0m, Why is t=1s?

Homework Equations

Vf = Vi + at
xf = xi + .5at^2

The Attempt at a Solution

xf = xi + .5at^2
0 = .5h + .5(-9.8)(1)
h = 9.8m

I know that this answer is not correct, I am having trouble figuring out how to manipulate the given data into one of those equations to either come up with h or find the total time.

all you need in this problem is
[tex] x_f = x_i + \frac{1}{2}gt^2[/tex]

A good place to start would be to find h in terms of t.

Also, the object travels .h in the last second which means that the displacement from when the ball hits the ground minus the displacement of where the ball was 1 second ago equals .5h.

 

[tiny-tim]

Hi CandyApples!

(try using the X² and X₂ tags just above the Reply box )

An object falls from height h from rest. If it travels .5h in the last second before hitting the ground, determine how long it takes to fall and the height h.
…
xf = xi + .5at^2

Nooo … learn your constant acceleration equations …

it should be x_f = x_i + v_it + .5at²

(and for that reason, you'll see it's not actually a very helpful equation to use! )

Hint: don't try to do it all in one equation … use two equations, both with v_i = 0.

 

[CandyApples]

Ok so trying a different method, I think I am a little closer.

Using the equation V_f² = 2ax I have determined that V_f = (19.6h)^1/2. The problem is that when I use this value for V_f=V_i+at I have two variables (h and t). Is there any way I can find t?

 

[tiny-tim]

Ok so trying a different method, I think I am a little closer.

Using the equation V_f² = 2ax I have determined that V_f = (19.6h)^1/2. The problem is that when I use this value for V_f=V_i+at I have two variables (h and t). Is there any way I can find t?

Nooo … why have you introduced v_f? … you're not asked for it, it only gives you an unwanted extra unknown

Hint: think of it as two projectiles …

you drop them both together from height h, one hits a ledge at height h/2, and the other hits the ground at height 0 one second later.

So for both of them, you have v_i a and s, and you have a relationship between t.

What do you get? ​

 

[CandyApples]

well if i treat them as two separate objects, s = V_it + .5at² will work to correlate the two by setting one t to either t+1 or t-1, right?

This will work if it is ok to set V_i equal to zero for both equations.

Equation 1 will be h = .5(-9.8)t²
Equation 2 will be .5h = .5(-9.8)(t-1)²

If this is logically set up, it will provide a way to link the two times together to find a numeric answer for h.

 

[rl.bhat]

well if i treat them as two separate objects, s = V_it + .5at² will work to correlate the two by setting one t to either t+1 or t-1, right?

This will work if it is ok to set V_i equal to zero for both equations.

Equation 1 will be h = .5(-9.8)t²
Equation 2 will be .5h = .5(-9.8)(t-1)²

If this is logically set up, it will provide a way to link the two times together to find a numeric answer for h.

Now substitute the value of h from the first equation in the second and solve the quadratic to find t. From that value you can find h.

 

[CandyApples]

Wow I got it now, thank you so much :-). I am so slow when it comes to physics.