Answer:
h=4+85*time - 1/2 32*time^2
16time^2-85*time-106=0
use the quadratic formula to solve for time (notice two solutions, one going up, one going back down).
time=(85+-sqrt(85^2+4*16*106) )/32
Explanation:
Recheck my work just in case I'm not good at this type of class but I wanted to help.
a. At maximum height h, the baseball has zero vertical velocity, so that with initial velocity v we have
[tex]0^2-v^2 = -2gh \implies h = \dfrac{v^2}{2g}[/tex]
Given that [tex]v = 30 \frac{\rm m}{\rm s}[/tex], the ball reaches a height of
[tex]h = \dfrac{\left(30\frac{\rm m}{\rm s}\right)^2}{2g} \approx \boxed{46\,\mathrm m}[/tex]
b. At max height, the baseball's vertical velocity is such that
[tex]0=v-gt[/tex]
For [tex]v = 30 \frac{\rm m}{\rm s}[/tex], we find
[tex]gt=30\dfrac{\rm m}{\rm s} \implies t = \dfrac{30\frac{\rm m}{\rm s}}g \implies t \approx 3.1 s[/tex]
which is half the time it spends in the air, making the total time about 6.1 s.
c. We saw in part (a) that
[tex]h = \dfrac{v^2}{2g}[/tex]
If we halve the initial speed and replace [tex]v[/tex] with [tex]\frac v2[/tex], we get
[tex]h = \dfrac{\left(\frac v2\right)^2}{2g} = \dfrac14 \cdot \dfrac{v^2}{2g}[/tex]
which means the max height is reduced by a factor of 1/4.
In part (b), we found the time to max height is
[tex]t = \dfrac{v}{g}[/tex]
and halving the initial speed gives
[tex]t = \dfrac{\frac v2}g = \dfrac12 \cdot \dfrac{v}{g}[/tex]
That is, the time to max height is reduced by a factor of 2 when the speed is halved, and so the overall time in the air is reduced by a factor of 2.
