Finding the height of a building

[moy13]

Homework Statement

Spiderman steps from the top of a tall building. He falls freely from rest to the ground a distance of h. He falls a distance of h/4 in the last interval of time of 1.2 s of his fall.

What is the height h of the building?

The acceleration is constant -g.

Homework Equations

y(t) = y(0) + v(0)t - 1/2gt^2
v = v(0) - gt
v^2 = v(0)^2 - 2g(y-y(0))

The Attempt at a Solution

I found the average velocity of the final interval of time to be (h/4)m / 1.2s = (h/4.8)m/s.

I know the average velocity occurs at half the displacement; so, I used this velocity as the initial velocity for another displacement of h/8 (or half the distance of the final interval).

Then I tried using g = -9.8m/s^2, v(0) = -(h/4.8)m/s, y(0) = (h/8)m, and y = 0 to find the time spiderman takes to travel that distance, but I cannot do that without figuring out h, my original target.

Please help.

 

[LowlyPion]

Write out equations for what you know.

h = 1/2*g*T²
T is your total time.

0 = 1/4 h - V_1/4 *(1.3) - 1/2*g*(1.2)²
V_1/4 is the velocity at 1/4 h.

Since he started from rest:
V_1/4 = g*(T - 1.2)

Solve for T and that will give you h won't it?

 

[moy13]

I tried solving for T but I still got a wrong answer for h

 

[rl.bhat]

0 = 1/4 h - V1/4 *(1.3) - 1/2*g*(1.2)2
It should be
0 = 1/4 h - V1/4 *(1.2) - 1/2*g*(1.2)2

 

[moy13]

Saw that 1.3 too, but I changed it to 1.2 and it's still wrong. Maybe I'm not understanding the concept. How do I solve for T?

Is it T = sqrt( 2h/g )?

and v(1/4) = g * ( sqrt( 2h/g - 1.2 )?

then solve for h. It's what I did, but I got 254m which is wrong.

 

[rl.bhat]

Put h = 1/2*g*T^2 and V1/4 = g*( T-1.2) in the equation. You get
1/4*1/2*g*T^2 - g*(T-1.2)(1.2) - 1/2*g*(1.2)^2 = 0
Cancell g through out and solve for T.

 

[LowlyPion]

Oops. Yes the 1.3 was a definite typo. Thanks for pointing that out rl.bhat.

Solving the quadratic gives me a different number than 254.

I can only suggest taking care with the signs and the algebra.