Calculating Ball Speed for Vertical Ascent

[B18]

Homework Statement

With what speed (in feet/second) must a ball be thrown vertically upward in order to rise to a height of 85.0 ft, neglecting air resistance?

Homework Equations

g=9.8 m/(s^2)
speed=distance/Δt

The Attempt at a Solution

I started off by setting speed=85.0ft/Δt
I understand I need to find the time duration.
I converted 85 ft to 25.908m.
Now I was thinking of multiplying 9.8 m/(s^2) by (1/25.908m) but that leaves me with seconds squared.

Am I missing a general formula for this problem? It seems like I don't have enough information at this point.

Thank you to anyone who can offer me any suggestions.

 

[nasu]

You need the equations for uniform accelerated motion.
speed=distance/t works only for uniform motion, motion with constant speed.

 

[B18]

So if i were to use a kinematic equation such as X(f)=V(ave)t+x(i) to find t how can I solve for t without knowing the velocity?

 

[CWatters]

Try one of the SUVAT equations for motion subject to constant acceleration..

http://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations

When I was at school it was worth memorising them as they help with all sorts of problems of this general type. Some 35 years later I can still remember equations 1,2 and 4.

 

[nasu]

You cannot.

But there is another kinematic equation.

 

[B18]

X(f)=X(i)+V(i)t+(1/2)a(t^2)
Im going to say this one with initial velocity set to 0 because in this problem that is the case.

 

[B18]

Figured it out. Used the kinematic equation above and set a=9.8m/s^2 thank you.

 

[CWatters]

X(f)=X(i)+V(i)t+(1/2)a(t^2)
Im going to say this one with initial velocity set to 0 because in this problem that is the case.

just for completness.. The final velocity will be zero not the initial velocity.

 

[B18]

So to ensure I understand this.. How does this look CWatters
Can I use V(f)^2=V(i)^2+2a(Xf-Xi)
solve for V(i)^2
V(i)^2=0-2(-9.8)(25.91)=22.54m/s
So in this problem a would be equal to -9.8 not positive??

 

[CWatters]

Yes. The initial velocity is in the opposite direction to the acceleration due to gravity - so velocity and acceleration will have different signs.