How Far Was the Second Student from the First When They Started Playing Catch?

[JimRaynor]

Homework Statement

2. Two students are playing catch. The first student throws the ball with an initial speed of 25.0 m/s at an angle of 30.0° above horizontal. The ball is 1.20 meters above the ground when it leaves her hand. The second student must accelerate at a rate of 0.750 m/s2 in order to catch the ball when it is 0.80 meters above the ground. How far away from the first student was the second student initially
if she started to run at the same time that the first student let go of the ball?

Homework Equations

x = xo + v(o)t
y = yo + .5(Voy + Vy)t
y = yo + Voyt + .5at^2
Vy = Voy + ayt
Vy^2 = Voy^2 + 2a(y-yo)

The Attempt at a Solution

I drew a picture of what I have and don't have, and i know the Vox and Voy, but I am lost at how to apply the 7.50m/s^2 accleration into the problem. I am pretty lost atm, and some help is appreciated

 

[BigJDubs]

. To begin, the maximum height of the ball is given by: y = yo + .5(Voy + Vy)t y = 1.2 + .5(25cos30)*t Therefore, the maximum height of the ball is given by: y = 1.2 + 12.5t Now, since we know that the ball is 0.8 meters above the ground when it is caught, we can set y = 0.8 and solve for t: 0.8 = 1.2 + 12.5t t = 0.32 seconds Now, we know the time of flight for the ball, so we can use the equation x = xo + v(o)t to find the horizontal range of the ball: x = xo + (25sin30)*t x = 0 + (25sin30)*0.32 x = 8 m Now, we have to take into account the acceleration of the catcher. We can use the equation y = yo + Voyt + .5at^2 to find the vertical position of the catcher at the time the ball is caught: y = yo + Voyt + .5at^2 y = 0 + 0 + .5(0.75)(0.32)^2 y = 0.12 m Now, we have to use the equation Vy^2 = Voy^2 + 2a(y-yo) to find the final velocity of the catcher: Vy^2 = 0 + 2(0.75)(0.12) Vy = 1.08 m/s Finally, we can use the equation x = xo + v(o)t to find the range of the catcher: x = xo + (1.08)*t x = 0 + (1.08)(0.32) x = 0.35 m Therefore,