How Does Velocity Change With Height in Projectile Motion?

[chessismyfort]

Homework Statement

A ball is thrown straight upward and rises to a maximum height of y above its launch point. Show that the velocity of the ball has decreased to a factor α of its initial value, when the height is y2 above its launch point is given by y2=(1-α^2)y.
Also SHOW that y2=(1-α^2)y.

Homework Equations

y=vt+1/2at^2, where a=-9.8m/s^2
v1=v0+at (maybe)
y2=(1-α^2)y

The Attempt at a Solution

I've already attempted a solution that involves plugging in y=vt +1/2at^2 and did a bunch of algebra; I was writing so fast I probably did something illegal; one thing i was worried about was plugging in (v1-v0)/t for the acceleration, and canceled out the t^2. I also attempted a solution that involved looking at derivatives of both sides. That is, y2' = ((1-α^2)y)' with respect to t, after plugging in the aforementioned equation of kinematics.

After all the algebra was said and done, I ended up with .5v(final)t+.5v(initial)t=y2, which definitely sounds incorrect to me.
Is there a way to modify the kinematics equation y=vt+1/2at^2 to include y2?

 

[ehild]

Homework Equations

y=vt+1/2at^2, where a=-9.8m/s^2
v1=v0+at (maybe)
y2=(1-α^2)y

Be careful with the notations. If you use y for the height in terms of t denote the maximum height by something else, say y_m. Also, vt in the equation for y(t) should be v₀t. So the correct form of the first equation is y=v₀t+1/2*at^2.
The equation for the velocity as function of time is correct in the form v=v₀t + at.

Plug in a=-g for the acceleration and work further with the new equations.

You have to prove that when the velocity is v=αv₀ the height is y=y_m(1-α^2).

... one thing i was worried about was plugging in (v1-v0)/t for the acceleration, and canceled out the t^2.

That is almost all right! But you cancel out t if you plug in t=(v-v₀)/a for t in the first equation. Then you will get the height y in terms of v instead of t.

Can you proceed?


ehild