Question

An object is dropped from a tower, 1600 ft above the ground. The object's height above the ground t seconds after the fall is s(t)=1600-16(t^2). Determine the velocity and acceleration of the object the moment it reaches the ground.

Answer 1

You'll need the time that the object will reach the ground, which can be obtained by solving for s(t) = 0:
1600 - 16t^2 = 0
1600 = 16t^2
100 = t^2
t = 10
Once you have this time, you can plug it into the velocity and acceleration function, which the first and second derivative of s(t), respectively.
v(t) = s'(t) = -32t
a(t) = -32 (as expected, free fall acceleration due to gravity is constant)
Then v(10) = -320 ft/s
And a(10) = -32 ft/s^2

Answer 2

First determine how long it takes for the object to hit the ground.
Set s(t) =1600 - 16t^2 = 0
So t = 10s would be the time taken.
Take the first derivative of s(t) to find the velocity-time relationship:
s'(t) = -32t
So s(10) = -320ft/s... This is your velocity.
I think the acceleration at all points close to the earth is roughly constant, -9.81m/s^2 = -32ft/s^2

Answer 3

thanks!

Answer 4

So velocity is first derivative?

Answer 5

yes