Answer
(a) After 2 seconds, the velocity of the rock is $7.56~m/s$ (upward)
(b) The velocity on the way up is $6.245~m/s$
The velocity on the way down is $-6.245~m/s$
Work Step by Step
(a) $h = 15t-1.86t^2$
$v = 15-3.72t$
After 2 seconds:
$v = 15-3.72(2) = 7.56$
After 2 seconds, the velocity of the rock is $7.56~m/s$ (upward)
(b) $h = 15t-1.86t^2 = 25$
$-1.86t^2+15t - 25 = 0$
$1.86t^2-15t + 25 = 0$
We can use the quadratic formula to find the times $t$ when the rock is 25 meters high:
$t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$t = \frac{-(-15) \pm \sqrt{(-15)^2-4(1.86)(25)}}{(2)(1.86)}$
$t = \frac{15 \pm \sqrt{225-186}}{3.72}$
$t = 2.3535~~$ or $~~t = 5.711$
We can find the velocity when the rock is 25 meters high on the way up:
$v = 15-3.72(2.3535) = 6.245~m/s$
We can find the velocity when the rock is 25 meters high on the way down:
$v = 15-3.72(5.711) = -6.245~m/s$
The velocity on the way up is $6.245~m/s$
The velocity on the way down is $-6.245~m/s$
