4. A football player passes the ball. The ball leaves his hand and is caught 1.7 seconds later by a receiver 30 meters away. (You may assume the ball was caught at the same height from which it was thrown.)

Question

a) What is the vertical component of the velocity of the ball just after it is released?
b) What is the speed (the magnitude of the velocity) of the ball when it is at the top of its trajectory?
c) What is the speed (the magnitude of the velocity) of the ball just before it is caught?

Answer 1

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