Expert: Sherry Wallin - 9/22/2009

Question
Question1
A projectile is launched at an angle from the horizontal. if air resistance is neglected then the formula for vertical displacement y of the projectile at time t after launch is given by y=(V(subscript)0sinA)t-1/2gt^2 where V (subscript)0 is the initial speed of the projectile and g is a constant called acceleration due to gravity. Derive expressions for
1) the vertical velocity of the projectile
2) the vertical acceleration of the projectile
and comment on the results  

Answer
Lorato~
    First let me tell you some standards for writing what you wrote that might make your life a little simpler. When you have a subscript just write is with an underscore in front of it so for v sub zero write v_0 so your equation becomes: y = (v_0sinA)t-(1/2)gt^2. Notice also the parenthesis around the 1/2, you need this because what you have written is equivalent to y = (v_0sinA)t-1/(2gt^2) which I am sure is not what you meant.

What you need to know to do this problem is that the first derivative of the displacement is the velocity and that the 2nd derivative of the displacement or the first derivative of the velocity is the acceleration. Let's check it out:

you are taking the first derivative with respect to t: y' = (v_0sinA)(1)-(2(1/2)gt = v_0sinA-gt
this is your velocity equation

the 2nd derivative y" = -g which is the acceleration of a projectile on earth, our gravity

I hope this helps.

Math Prof