Question

What is the minimum speed a ball must be thrown upward if it is to reach a height of 11 feet?

Answer 1

What part has you lost. Do you know the basic equations of motion for projectiles?

Answer 2

honestly I don't. This is my calc hw and I have never taken physics.

Answer 3

Ah okay.
Well let's call the position x(t), velocity v(t)=x'(t) and acceleration a(t)=x''(t)
We know that x''(t)=-g. So you can play around with this.
One approach would be to find the time when x'(t)=0 which is the time of maximum height. Then see if you can't get x(t_max) to evaluate to 11

Answer 4

-.5gt^2 = x(t) correct?

Answer 5

You are forgetting the initial consitiond -- the constants of integration shouldnt just disappear.
Commonplace is to do x(0)=x0 and x'(0)=v0

Answer 6

ya i am lost. sorry

Answer 7

I mean how did you get from x''(t)=-g to your x(t)?

Answer 8

the integral of -g = -gt+c and then the integral of -gt+c = -.5gt^2+ct+C. I assuming the c's are zero because i am starting at the origin

Answer 9

Well one of the cs will not be. When you go from x''(t)=-g to x'(t)=-gt+c you need to set x'(0) to the initial condition which we will just call v0 for now. So x'(0)=v0. So that c can be solved for. Generally you do the same thing for position by setting x(0)=x0

Answer 10

i will give that a shot

Answer 11

after attempting your approach i get a solution of 8sqrt(11). that look correct?

Answer 12

sorry all of our solutions have to be in square roots so some sort.

Answer 13

What did you get for x(t)?

Answer 14

after analysis the question a little further i realized that i only need to find the velocity so i asked a buddy and he gave me the equation of vf^2=v0^2+2ad which i transposed to sqrt(Vf-2AD) which came out to sqrt(0-2*32*11)

Answer 15

Well sure the physics approach is easier knowing the equations but I thought you wanted to do it with calc.

Answer 16

i do lol, my friend said this is easier approach and I was lost with the calc approach. this is just the 2 week of calc and it has been crazy since it is online.