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.)

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?

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

To solve this problem, we need to consider the vertical motion and the horizontal motion of the ball separately. Let's break it down step by step.

a) To find the vertical component of the velocity of the ball just after it is released, we need to determine the initial vertical velocity. We can use the equation:

v = v0 + at

Where:
v = final velocity (vertical component)
v0 = initial velocity (vertical component)
a = acceleration (in this case, the acceleration due to gravity, which is approximately -9.8 m/s^2)
t = time (1.7 seconds, as given in the problem)

Since the ball is caught at the same height that it was thrown, the final vertical velocity will be zero. Therefore, the equation becomes:

0 = v0 + (-9.8 m/s^2)(1.7 s)

Now we can solve for v0:

v0 = (9.8 m/s^2)(1.7 s) = 16.66 m/s^2 (upward)

So, the vertical component of the velocity of the ball just after it is released is 16.66 m/s^2 upward.

b) To find the speed (the magnitude of the velocity) of the ball when it is at the top of its trajectory, we need to consider the vertical motion only. At the top of its trajectory, the vertical velocity will be zero. We can use the equation:

v^2 = v0^2 + 2aΔy

Where:
v = final velocity (vertical component)
v0 = initial velocity (vertical component)
a = acceleration (in this case, the acceleration due to gravity, which is approximately -9.8 m/s^2)
Δy = change in vertical position (in this case, the maximum height reached by the ball, which is unknown)

At the top, the final velocity v will be zero. Therefore, the equation becomes:

0 = v0^2 + 2aΔy

We can solve for v0:

v0^2 = -2aΔy
v0 = √(2aΔy)

Since the ball reaches its maximum height, Δy, and then comes back down to the height it was thrown, Δy = 0. Therefore:

v0 = √(2a(0))
v0 = 0

So, the speed of the ball when it is at the top of its trajectory is zero. This makes sense, as the ball momentarily stops before changing direction.

c) To find the speed (the magnitude of the velocity) of the ball just before it is caught, we need to consider the horizontal motion only. We know that the horizontal distance traveled by the ball is 30 meters. We can use the equation:

x = v0t + (1/2)at^2

Where:
x = horizontal distance (30 meters)
v0 = initial horizontal velocity (which is equal to the final horizontal velocity since there is no horizontal acceleration)
t = time (1.7 seconds)

Since there is no horizontal acceleration, the equation becomes:

x = v0t

We can solve for v0:

v0 = x/t
v0 = 30 m / 1.7 s
v0 ≈ 17.65 m/s

So, the speed of the ball just before it is caught is approximately 17.65 m/s.