If a rock is propelled upward from the surface of Mars at a velocity of 93 m/sec, it reaches a height

s = 93t – 1.86t^2 after t sec.
a. Find the rock’s velocity and acceleration at time t.

v'(t)=93-3.72t
a''(t)=-3.73

b. How long does it take the rock to reach its highest point?

v'(t)= 93-3.72t=0

t=25s
c. How high does the rock go?

f(x)= 93(25)-1.86(25)^2=1162.5m

I couldn't find the last two.

d. How long does it take the rock to reach 1/4 of its maximum height?
e. How long is the rock aloft?

a, b, and c look correct. For d, divide your answer from c by 4, plug it into the height equation, and solve for t. For e, it took 25 to reach the peak so it will take twice that long to reach 0 again (set s=0 and solve for t).

To find the answers to parts (d) and (e), we will use the given equation for the height of the rock, which is s = 93t – 1.86t^2.

d. To find how long it takes for the rock to reach 1/4 of its maximum height, we need to find the time when the height is equal to 1/4 of the maximum height. Let's call this time T.

We know that the maximum height occurs at the highest point of the rock's trajectory, so the highest point is when the velocity of the rock is zero. We found in part (b) that the velocity equation is v'(t) = 93 - 3.72t. Setting this equation equal to zero and solving for t, we found t = 25s.

Now, we need to find the value of the height at t = 25s, which will give us the maximum height. Plugging this value into the equation for height:
s = 93 * 25 - 1.86 * 25^2 = 1162.5m

Now we have the maximum height, which is 1162.5m. To find 1/4 of this height, we can simply multiply the maximum height by 1/4:
1/4 * 1162.5 = 290.625m

Now we set up the equation s = 290.625 and solve for t:

290.625 = 93t - 1.86t^2

This is a quadratic equation, so we can solve it by setting it equal to zero:
1.86t^2 - 93t + 290.625 = 0

Solving this quadratic equation will give us two values for t - one for when the rock is on its way up, and the other for when it's on its way down. We can then find the positive solution, which will represent the time it takes for the rock to reach 1/4 of its maximum height.

e. To find how long the rock is aloft, we need to determine the time it takes for the rock to hit the ground again. Since the height of the rock is zero when it hits the ground, we can set the height equation equal to zero and solve for t:

0 = 93t - 1.86t^2

This is again a quadratic equation, so we solve it by setting it equal to zero:
1.86t^2 - 93t = 0

Solving this equation will give us the two values for t. We can then find the positive solution, which will represent the time the rock is aloft, or in other words, the time it takes for the rock to hit the ground again.

By solving these equations, you will be able to find the answers to parts (d) and (e).