Finding drop height h when ball falls 0.47h in final second

[mathphysicnb]

Homework Statement

A ball is dropped from a height h. The ball displaces 0.47h meters from the time it hits one second before it drops to the ground and when it hits the ground. Find h.

Homework Equations

xf-xo=volt + 0.5gt^2

The Attempt at a Solution

h = ?
t1(at 1 second before ball hits ground) = 1
s(at one second before till it hits ground) = 0.47h

So, this is what it visually looks like:

o (A) t=0, v=0, a=g, h=0
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| v
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| (B) t=t1, a=g, h=0.53h
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| (C) t=t2, a=g, h=H

So, from A to B, the ball falls 0.53h meters. From A to C, the ball falls H meters. And From B to C, the ball has fallen 0.47h meters in 1 second, which comes from the problem statement. So the ball at B has fallen t seconds. And when the ball reaches C, it has fallen t + 1 seconds.

I need to find h...

So, I use the formula above for A to B and again for A to C:

0.53h = 0.5gt^2 => h= (0.5g/0.53)t^2 (A to B)

h = 0.5g(t+1)^2 (A to C)

I have set both equations equal to h. Now I solve for t by setting said equations equal to each other:

(0.5g/0.53)t^2 = 0.5g(t +1)^2 =>

t^2/0.53 = (t+1)^2 =>

0 = (1-(1/0.53))t^2 + 2t + 1

Using quadratic formula, I get approximately 2.68 seconds.

I plug this back into one of the formulas above to get h:

h = 0.5g(t+1)^2 =~ 66.43 meters.

Is this correct?

I have to enter this online and I literally have one more attempt left before whatever answer I put gets submitted. Please help me and let me know if this is correct.

Thank you.

 

[rl.bhat]

Your answer is correct.