Answer:
a. To find the velocity and acceleration of the rock at time t, we need to take the first and second derivatives of the height equation with respect to time:
s = 28t - 0.8t^2
v = ds/dt = 28 - 1.6t
a = dv/dt = -1.6
So the velocity of the rock at time t is v = 28 - 1.6t m/s, and its acceleration is a = -1.6 m/s^2.
b. To find how long it takes the rock to reach its highest point, we need to find the time at which the velocity is zero. We can set v = 28 - 1.6t = 0 and solve for t:
28 - 1.6t = 0
t = 17.5 seconds
So it takes 17.5 seconds for the rock to reach its highest point.
c. To find the maximum height reached by the rock, we can substitute t = 17.5 seconds into the height equation:
s = 28t - 0.8t^2
s = 28(17.5) - 0.8(17.5)^2
s = 245 meters
So the rock reaches a height of 245 meters.
d. To find how long it takes the rock to reach half its maximum height, we can set s = 245/2 = 122.5 meters and solve for t:
s = 28t - 0.8t^2
0.8t^2 - 28t + 122.5 = 0
t = 7.5 seconds or t = 20 seconds
So it takes the rock 7.5 seconds to reach half its maximum height.
e. To find how long the rock is aloft, we need to find the time at which it returns to the surface. We can set s = 0 and solve for t:
s = 28t - 0.8t^2
0 = 28t - 0.8t^2
t = 35 seconds
So the rock is aloft for 35 seconds.