Thrown Upwards with one variable

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A stone is thrown vertically upward, passing point A at speed v and point B, 3.00 m higher, at speed 1/2 v. To find the initial speed v, the kinematic equation V^2 - Vo^2 = 2as is suggested, with appropriate substitutions for velocities and acceleration due to gravity. The maximum height above point B can be calculated using the same kinematic equation, setting the final velocity at the peak to zero. Careful attention to the signs of the variables is crucial, as upward motion is considered positive while gravity acts negatively. This approach will yield both the initial speed and the maximum height reached by the stone.
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Homework Statement



A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 3.00 m higher than A, with speed 1/2 v. Calculate (a) the speed v and (b) the maximum height reached by the stone above point B.

Homework Equations


Time in the sky= Initial Velocity/9.8m/s^2
Maximum Height= xmax = Initial Velocity^2/ 2 x (9.8/m/s^2)

The Attempt at a Solution



I've been experimenting and can't seem to find an answer. I've tried using x+3+y to fill in for the maximum height, but I'll always end up more than one variable to solve for. Because deceleration would be at gravity, 9.8m/s^2, I was trying to figure out a way to even calculate a time, but without an initial velocity, it seems hopeless.

Thanks for the help :)
 
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first check your relevant equations... they don't look right to me although they are written in some weird notation. check out the kinematics equations for constant acceleration at say
http://en.wikipedia.org/wiki/Kinematics
should be a good start (although no one should really take wiki seriously :smile:)
 
oh.. and of course, welcome to PF.
 
Thanks for the warm welcome, and I'll be sure to look it over :)
 
In all such problems be careful of the sign (+ or -) of different quantities. First decide the +ve direction for displacement. If the object is thrown upwards, it is logical to have upward displacement as +ve. However, if the object is thrown upwards from top of a tower, there is an option to have upward or downward displacement as +ve. Then, the velocities and accelerations in the +ve direction are also +ve. Acceleration due to gravity is always downwards. Hence, it will be +ve or -ve depending upon the direction chosen as positive. In the present case it will be -ve as upwards is +ve.

Consider motion of the stone from point A to point B. Use equation V^2 - Vo^2 = 2as and substitute Vo = V, V = V/2, s = 3m and a = - 9.8 m/s^2 and solve for V.

To get the max height reached above B, again use the above equation with substitution : V = 0, Vo = V/2 (for which now we have a value) and a = -9.8 m/s^2 and solve for s.
 
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