Question

Calculus Derivatives : a projectile is fired vertically upward with an initial velocity of 16 ft/sec.
1)How long will it take for the projectile to hit the ground?
2) how high will it go?
I thought it would use the equation S(0)+V(0)t- (1/2)gt^2 but I'm not sure what to do at the moment.
S(0) Initial position
V(0) initial velocity
and g = gravity

Answer 1

S(t)= S(0)+V(0)t- (1/2)gt^2

put in the numbers
\[ S(t)= 0 + 16 t - \frac{1}{2} 32 t^2 \\
S(t)= - 16 t^2+16 t \]
this is a parabola, so you can find its vertex (t= -b/(2a) )
or use calculus to find where the slope is 0
you can find the total flight time by solving for S(t)=0
or you can double the time it takes to reach the peak (it takes as much time to go up as down)

Answer 2

Phi, I thought you had it right the first time. Can you explain to me where the 32 came from for g?

Answer 3

Since S(T)= -16t^2+16t that counts as its position.
and then Velocity would be = -32t+16
Acceleration would be -32.
Wouldn't setting the velocity to 0, would give me the time as when its at its highest peak, showing me the largest height? 0=-32t+16 = t= -2. And then putting it into S(T)
-16(-2)^2 + 16(-2) Would lead me to when its still at its highest point showing me how high it will go?

Answer 4

@jb1515g gravity I believe is 32.174 ft/s2 so thats where he got 32

Answer 5

Ah, I see now. I didn't notice it was in feet. My mistake. I was working in meters.

Answer 6

yes, I used meters/sec^2 for g in the first post, and then noticed the problem is using feet for units

Answer 7

Yeah, I did the same thing, just didn't notice it.

Answer 8

Would setting S(T) = 0 = -16t^2+16t equal to how long it will take for the projectile to hit the ground?

Answer 9

Check your math, marsxtc. In 0=-32t+16, t isn't -2.

Answer 10

Ah I see, I messed up a bit I wast trying to do it fast. So it'd be 1/2.

Answer 11

No, taking S(T) = -16t^2 + 16t and plugging in t will give you the max height. To find how long it would take to hit the ground, you take the t you just found (which is the time it takes to get to the max height) and double it, since it will take the same time to fall from that height.

*edit for clarification

Answer 12

***
Would setting S(T) = 0 = -16t^2+16t equal to how long it will take for the projectile to hit the ground?
***

or solving for t when the height is 0 (i.e. on the ground) we find
\[ t (-16t + 16) = 0 \\
t= 0 \text{ or } -16t+16=0 \\ -16t= -16 \\ t = 1 \]
so t=0 (at launch) and t= 1 (back on the ground)

Answer 13

So the time is 1/2 for it to reach maximum, so 1 second for it to reach the ground?

Answer 14

Yes sir.

Answer 15

So according to phi plugging in 0 as the position would also work wouldn't it?

Answer 16

Yeah, that works too. I always found finding for max height, then doubling, easier, But both work.

Answer 17

Ok just to clarify ( I have a huge test on derivatives and I forgot how to do everything related to this if theres no like equation given to me) Finding the S(T), I can determine the V(T) and the A(T). Setting V(T) to 0 Will give me the time for the maximum height because at velocity = 0 its yeah isn't moving. inputting that time into S(T) will give me the height? And setting S(T) to 0 will give me the time of when it hits the ground?

Answer 18

Yes, that is correct.

Answer 19

Alright. I really don't know on who to give the medal to. Both of you were so helpful!...

Answer 20

I'd say Phi is deserving

Answer 21

Alright thank you for taking your time to help me!