Thrown Upwards with one variable

[sylenteck0]

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 :)

 

[mjsd]

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 )

 

[mjsd]

oh.. and of course, welcome to PF.

 

[sylenteck0]

Thanks for the warm welcome, and I'll be sure to look it over :)

 

[Vijay Bhatnagar]

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.