A ball thrown straight up from the ground passes a window 5.6 m up. An observer looking out the window sees the ball pass the window again, going down, 3.4s later.

Find the velocity with which the ball was initially thrown.

Use:

s(t) = s(0) + v(0) t - ½ g t²
v(t) = v(0) - g t
9 = 9.8[m/s²]

Let the ball pass the window up at t=0[s] and down at t=3.4[s], so:
v(3.4) = -v(0)
v(3.4) = v(0) - 9.8×3.4[m/s]
.: v(0) = 9.8×3.4/2 [m/s]
v(0) = 16.66[m/s]

This is the velocity at the window sill.
Let the ball have been thrown at t = ŧ.
s(ŧ) = -5.3[m/s]
s(ŧ) = 16.66 ŧ - ½×9.8 ŧ²
.: 4.9 ŧ² -16.66 ŧ - 5.3 = 0
ŧ = (16.66±√(16.66²+4×4.9×5.3))/(2×4.9)
ŧ = 1.7±1.993...[s]
The negative root gives the time of throw, while the positive gives the time of return.
.: ŧ ≈ -0.293...[s]

Initial velocity at time of throw:
v(ŧ) = v(0) - g ŧ
.: v(ŧ) = 1.66 + 9.8 × 0.293...
v(ŧ) = 6.57...[m/s]

To find the velocity with which the ball was initially thrown, we can use the kinematic equations of motion.

Let's define the following variables:
v_i: Initial velocity of the ball (what we need to find)
v_f: Final velocity of the ball (when it reaches the window going down)
s: Distance traveled by the ball (5.6 m)
t: Time it takes for the ball to reach the window going up

Since the ball passes the window going up and then again going down, the total time for these two events is 3.4 seconds.

The first thing we need to do is find the time it takes for the ball to reach the window going up (t). We can use the formula:

s = v_i * t + 0.5 * g * t^2

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Plugging in the values we know:
5.6 = v_i * t + 0.5 * 9.8 * t^2

Now, we can use another equation to represent the total time for the two events:

t + (t + 3.4) = total time = 2 * t + 3.4

Simplifying the equation:
2t + 3.4 = 3.4

Solving this equation:
2t = 0
t = 0

This implies that it takes 0 seconds for the ball to reach the window going up, which means the ball was already at the window when it was thrown up. This tells us that the initial velocity of the ball (v_i) is 0.

Thus, the velocity with which the ball was initially thrown is 0 m/s.

To find the velocity with which the ball was initially thrown, we need to break down the problem into two parts: the upward journey and the downward journey of the ball.

Step 1: Determine the time taken for the ball to reach the window on its way up.
From the given information, we know that the ball passes the window again, going down, 3.4 seconds after passing it on the way up. This means the total time taken for the complete journey (up and down) is 3.4 seconds.

Step 2: Find the time taken for the ball to reach the highest point.
Since the ball takes the same amount of time to go up and come down, half of the total time is taken for the upward journey. Therefore, the time taken for the ball to reach the highest point is 3.4 seconds divided by 2, which is 1.7 seconds.

Step 3: Calculate the velocity of the ball on its way up.
To calculate the velocity of the ball on its way up, we can use the equation:

v = u + at

where:
v = final velocity (which is 0 m/s at the highest point)
u = initial velocity
a = acceleration (in this case, acceleration due to gravity, -9.8 m/s^2)
t = time taken (which is 1.7 seconds)

Since the final velocity is 0, we can rearrange the equation to solve for the initial velocity:

0 = u + (-9.8) * 1.7

Rearranging the equation gives us:

9.8 * 1.7 = u

u = 16.66 m/s (rounded to two decimal places)

So, the velocity with which the ball was initially thrown is approximately 16.66 m/s.