Simple throwing ball up problem

[j_suder2]

Simple throwing ball "up" problem

Homework Statement

Ball is thrown vertically with upward velocity of 18m/s when it reaches 1/4 of its maximum height above its launch point. Find initial launch velocity of ball.

Homework Equations

0.5mv2, mgh, v=d/t

The Attempt at a Solution

The 1/4 of the maximum height is throwing me off.
I tried using kinematics equations to first solve for the first part of the throw (the 1/4)
and then use final velocity of zero at the max height using deceleration of gravity to find this height.
I'm stumped, any help is appreciated.
Thanks.

 

[Dale]

I would do this with conservation of energy. Set your h=0 point at the 1/4 point (where v=18m/s)

 

[robphy]

v is not generally equal to d/t.
While you can use energy conservation, in case you haven't gotten that far... or wish to use something more kinematical, you can use a formula (resembling the energy conservation approach) derived from the constant-acceleration position-vs-time and velocity-vs-time equations.

 

[j_suder2]

Ok thanks, I found the height to be 32.8 m (?) at the quarter point, then set KE + PE final to = KE initial and got 25 m/s for an initial velocity. The answer is 21 m/s though. Not sure where I went wrong.

 

[kplooksafterme]

Your height value is incorrect. I'd use conservation of energy to find the height (you don't need to set the height at the 1/4 mark to zero, I find this adds an unnecessary step; just set it equal to 1/4h). What is your energy conservation expression? Once you have the correct value of the height, it's merely a kinematics problem.

 

[rl.bhat]

If h is the maximum height, you can wright two equation.
v1^2 - vi^2 = -2gh/4...(1)
0 - vi^2 =-2gh ...(2) solve these equation and find vi

 

[kplooksafterme]

rl.bhat, those equations are wrong... should be v1^2-vi^2 = -2gh/4...

 

[j_suder2]

My conseravtion of energy expression is .5mv^2 + mgh = .5mv^2
This leaves me with two variables, so I am not sure what is next.
Thanks.

 

[Hootenanny]

My conseravtion of energy expression is .5mv^2 + mgh = .5mv^2
This leaves me with two variables, so I am not sure what is next.
Thanks.

Perhaps this would be helpful;

Ball is thrown vertically with upward velocity of 18m/s when it reaches 1/4 of its maximum height above its launch point. Find initial launch velocity of ball.

 

[kplooksafterme]

Set it up so your initial velocity is 18m/s. What is the height when the velocity is 18 m/s? What is the velocity when the ball is at maximum height? Once you know the height, the rest will fall into place...