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.
xf-xo=vot + 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
| (B) t=t1, a=g, h=0.53h
| (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.