Question

Question on derivatives and antiderivatives. Question attached.

Answer 1

\[h(t)=-9.8t ^{2}+50\]
Since the acceleration is negative and 9.8 it will be the leading coeffecient, because velocity starts at rest, that term (0x) is omitted, and the starting position is 50 meters

Answer 2

so is 50 the asnwer to part B? And also how do we find the velocity in part C and part D?

Answer 3

To find the answer to part B you set h(t) = 0 and solve for t...
For part c you must derive the equation to find velocity!

Answer 4

HOW do we derive the eqaution? Are we suposed to antidiff the h(t)

Answer 5

im confused what u mean by derive the equation

Answer 6

Another term is to....find the instaneous rate of change of the function h(t), in this case it would be...\[v(t)=-19.6t\]

Answer 7

ok. So now do we set v(t) equal to 0 too to find part C

Answer 8

Using the time you found in the previous problem, plug it in to the velocity equation!

Answer 9

Ok so I got t=2.26 and so v(t) = -44.296. So now for D which equation do we plug 5 m/s into?

Answer 10

do we set 5 equal to h(t) amd solve for t

Answer 11

Okay, so saying it has an initial velocity downward of 5 would change the h(t) equation.....\[h(t)=-9.8t ^{2}-5t+50\]
Again...solve for t!

Answer 12

SO do i set that equal to 0 and solve for t?

Answer 13

Yes! It should be a smaller time than before, which makes sense, if you throw it downwards it should arrive at the ground even quicker than before!

Answer 14

h(t) was actually -4.9t^2+50 but its ok. After that i jsut used that answer to do the rest.

Answer 15

OH! I'm so sorry! Yes it was supposed to be halved!