Posted by Anonymous on Friday, October 5, 2012 at 1:14am.
vx(t) = vx0
vy(t) = vy0 - g*t
y = vy0*t - 1/2*g*t^2
x = vx0*t
where y is vertical distance measured from the height of the football player's hand, x is the horizontal distance, t is time, vy0 is the initial speed in the y direction, vx0 is the initial speed in the x direction, and vy(t) is the speed in the y direction as a function of time, and vx(t) is the speed in the x direction as a function of time
When the football player catches the ball at 1.7s, the vertical height y is back at 0, and the horizontal distance is at 30
0 = vy0*1.7 - 1/2*g*1.7^2
30 = vx0*1.7
Solve for vy0 and vx0
a) The vertical component of the velocity of the ball just after it is released is vy(t) evaluated at t = 0
vy(t) = vy0
b) At the top of its trajectory, the ball will be exactly halfway to the receiver, so t = 1.7/2 = 0.85 s. At this time
vy(t) = vy0 - g*0.85
vx(t) = vx0
solve
then the magnitude of the speed is
(vy(t)^2 + vx(t)^2)^0.5
c) Just before the ball is caught, t = 1.7s
vy(t) = vy0 - g*1.7
vx(t) = vx0
solve for vy(t) and vx(t), and then use the formula from part b to calculate the magnitude of the vector
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