This calculus stuff is really getting me any help is greatly appreciated!!?

Question

On the moon the acceleration due to gravity is 5ft/sec^2. An astronaut jumps into the air with an intitial velocity of 10 ft/sec. How high does he go? How long is it before he comes back down?

Answer 1

His vertical acceleration is given by:

a = -5

right? I have made it negative because i have designated downwards as negative. Integrate wrt. time to get velocity:

v = -5t + C

We know that the initial velocity is 10 (upwards, positive), so the velocity is given by:

v = -5t + 10

Integrate once more to get displacement as a function of time:

x = -2.5t^2 + 10t + D

He is initially on the ground, so x(0) = 0, i.e. D=0.

x = -2.5t^2 + 10t

Now, he is at the maximum height when the velocity is zero. The velocity is given by:

v = -5t + 10

Set it equal to zero and solve for t:

-5t + 10 = 0
Therefore t = 2

So at t=2, he reaches maximum height. Plug t=2 into the displacemenet equation to find what his max height is:

x = -2.5t^2 + 10t
x(2) = 10

Therefore, he reaches max height of 10ft at t=2 seconds.

To find how long he is in the air for, set the displacement equal to zero i.e. solve for t when his displacement is zero again:

0 = -2.5t^2 + 10t
t = 0, t=4

Now t=0 should be obvious, since he starts from the floor. t=4 means that after 4 seconds his displacement is zero again i.e. he is back on the ground.

Answer 2

you have to use the classic f(x) = -.5gt^2 + vt + s

in this case, g = 5 ft/sec^2, v = 10 ft/sec and s = 0 since he is on the surface.

your function is f(x) = -2.5t^2 +10t

you can set this equal to zero to find the time spent -- in this case, t(-2.5t + 10) = 0, so t = 0 or t = 10/2.5 = 4 sec

he reaches the max height halfway between launch and landing, so this occurs at 2 sec

substitute 2 into the function to get f(2) = -2.5(2)^2 + 10(2)
f(t) = -10 + 20= 10 ft
so his max height is 10 ft

Answer 3

V = 10 - 5t

He will go up for 2 seconds then down for 2 seconds.

Integrate V to get distance to find he went up 10 ft.

Answer 4

v = u + at
0 = 10 - (5) t
1.524 t = 3.048
t = 2 s

time taken to reach bottom = 2 x 2 = 4 s