An object is dropped from a tower, 167
ft above the ground. the object's height above ground t sec into the fall is
s=167-16t^2
a. what is the object's velocity, speed, and acceleration at time t?
b. about how long does it take the object to hit the ground?
c. what is the object's velocity at the moment of impact?

A)
Differentiating the given equation
ds/dt=-32t
Which is the expression for velocity, hence the velocity of the object will be -32t at time t,
B)
When the object hits the ground, the distance above the ground I.e s equals 0, returning the value in given equation
0=167-16tsquare
T=3.23 seconds
C)
Returning the value from b) in to equation form a)
Ds/dt=-32t
V=-32*3.2
V=-102.4m/sec squares
In case you are wondering why the answer is negative, usually when velocity is positive, distance increases, but in this equation the distance I.e above the ground decreases with increase in velocity.

328.75

Step-by-step explanation:

First you multiply by 5 because quintuple=5

Then you will get 328.75

Hope this Helps!! :)