Question

An object is dropped from a tower, 163 ft above the ground. The object's height above ground t sec into the fall is s=163-16t^2.

a. The object's velocity at time t is ___.
b. The object's speed at time t is ___ ft/sec^2.
c. The object's acceleration at time t is ___ ft/sec^2.
d. It takes ___ sec for the object to hit the ground.
e. The object's velocity at the moment of impact is ___ ft/sec.

I need some simple explanations on how to do this and the answers if possible. Thank you! I know you have use the 1st/2nd derivative, please explain that as well if need be.

Answer 1

First find the derivative of s

Answer 2

@StudyGurl14 is it -32t?

Answer 3

yes

Answer 4

Okay, so that's the velocity

Answer 5

alright

Answer 6

speed is the absolute value of velocity

Answer 7

would it take 3.2 seconds for the object to hit the ground? if I round to the nearest tenth?

Answer 8

@StudyGurl14 i think i have everything besides the acceleration. do i have to use the change in velocity over time to find it?

Answer 9

accerlation is the derivative of velocity

Answer 10

and the 3.2 is right

Answer 11

so just -32 for the acceleration or?

Answer 12

yep

Answer 13

now you have e left

Answer 14

for part e, is it just 0? or does it mean the speed right before it hits the ground?

Answer 15

p[lug in the answer to part d for e

Answer 16

oh wait, yeah

Answer 17

for t i mean, to answer e

Answer 18

it comes out to 163-163.84?

Answer 19

no plug it into the velocity equaiton

Answer 20

so -102.4 ft/sec? do i leave it as a negative?

Answer 21

Yes. That's the velocity. 102.4 would be the speed

Answer 22

Nice job :)

Answer 23

thank you so much. i've only got 5 questions, that was the first one. i'll probably be asking about a couple more if i can't figure them out.

Answer 24

Ok, lol