Question

A tennis ball bounces so that its initial speed straight upwards is b feet per second.
Its height s in feet at time t seconds is given by s = bt − 16t
2
a) Find the velocity v = ds/dt at time t.
b) Find the time at which the height of the ball is at its maximum height.
c) Find the maximum height.
d) Make a graph of v and directly below it a graph of s as a function of time. Be sure
to mark the maximum of s and the beginning and end of the bounce.

Answer 1

e) Suppose that when the ball bounces a second time it rises to half the height of the
first bounce. Make a graph of s and of v of both bounces, labelling the important points.
(You will have to decide how long the second bounce lasts and the initial velocity at the
start of the bounce.)
f) If the ball continues to bounce, how long does it take before it stops

Answer 2

In part e) I tried approaching the problem by setting t=b/16+t to find out the time it takes to reach half the height reached by the ball in its second bounce. However, when I used the quadratic formula to solve for t, both solutions were negative and thus are rejected.

Answer 3

@DanJS

Answer 4

a)b-32t
b) t=b/32
c) b^2/64

Answer 5

Yeah looks good so far...

Answer 6

\[s = b*t-16*t^2\]
\[v=b-32*t\]
The max height for the above first bounce is
s = b^2/64 when t=b/32, half of that height would be b^2/128

When does that height happen in the first bounce t=?
s = b*t - 16t^2
\[\large s=\frac{ b^2 }{ 128 }=b*t-16*t^2\]

Answer 7

But won't b^2/128=bt-16t^2 give the times at which the ball is b^2/128 from the ground in it's first bounce.

Answer 8

The quadratic looks like it goes to..
\[16t^2-bt+\frac{ b^2 }{ 128 }=0\]
\[\large t=\frac{ b \pm \sqrt{b^2-4*16*\frac{ b^2 }{ 128 }} }{ 2*16 }=\frac{ b \pm \sqrt{b^2/2} }{ 32 }=\frac{ b \pm \frac{ \sqrt{2}*b }{ 2 } }{ 32 }\]
\[\large t=\frac{ b*(\sqrt{2}\pm2 )}{ 64 }\]

Answer 9

yeah, those are the times when it is at half the maximum height. You want the second bounce to have a maximum value of this height.

Answer 10

The ball's final velocity in its first bounce was -b and it would have attained the same height if we are assuming an ideal case where no external forces are acting on the ball such as friction. However, the ball rose to half the height of its first bounce. So should the equation describing the balls vertical position still hold?

Answer 11

yeah, the motion equations holds all the time, the only difference is the initial velocity of the second bounce.
The First bounce reaches half it's max height when \[t=\frac{ b*(2 \pm \sqrt{2}) }{ 64 }\]
At these times, the velocity is...

\[\large v = b - 32*t\]
\[v = b - 32*\frac{ b(2 \pm \sqrt{2}) }{ 64 } = \frac{\pm \sqrt{2} }{ 2 }*b\]

Answer 12

So with that, you have the initial velocity for the next bounce...root(2)/2*b, ill use th epositive for the initial is on the way up.
use that v now and put in place of b

s = \[\large s=\frac{ \sqrt{2} }{ 2 }*b*t -16t^2\]

Answer 13

The second bounce has equations..
\[\large s=\frac{ \sqrt{2} }{ 2 }*b*t-16*t^2\]
\[\large v=\frac{ \sqrt{2} }{ 2 }*b-32*t\]

Answer 14

Here are some graphs of the position and velocity of two bounces..first in red, second in black. and also the velocities in green and blue...

Answer 15

I just put a slider for the 'b' value, notice it is set at 100 currently for that graph..