To solve this problem, we need to analyze the motion of the ball using the equations of kinematics.
Let's denote:
- \( h = 7.5 , \text{m} \) (the height of the bottom of the window),
- \( d = 2 , \text{m} \) (the height of the window),
- \( H_t = h + d = 9.5 , \text{m} \) (the height of the top of the window),
- \( t = 1.3 , \text{s} \) (the time from passing the bottom of the window to the top of the window).
We know that the ball is thrown up, slows down under the influence of gravity, and eventually falls back down. The acceleration due to gravity (\( g \)) is approximately \( 9.81 , \text{m/s}^2 \).
Step 1: Finding the speed at the position of the bottom of the window
Using the kinematic equation: \[ s = v_i t + \frac{1}{2} a t^2 \] where \( s \) is the displacement, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
Let’s find the displacement for the ball traveling from the bottom of the window (\( 7.5 , \text{m} \)) to the top of the window (\( 9.5 , \text{m} \)). The displacement \( s \) here is: \[ s = H_t - h = 9.5 - 7.5 = 2 , \text{m} \]
When the ball moves from the bottom of the window to the top, it is moving upwards with time \( t \) of \( 1.3 , \text{s} \). We can write the equation for the upward motion: \[ 2 = v_b (1.3) - \frac{1}{2} (9.81)(1.3^2) \]
Where \( v_b \) is the velocity of the ball as it passes the bottom of the window.
Calculating the acceleration term: \[ \frac{1}{2} (9.81)(1.3^2) = \frac{1}{2} (9.81)(1.69) \approx 8.292 , \text{m} \]
Substituting this into the equation: \[ 2 = v_b (1.3) - 8.292 \] \[ v_b (1.3) = 2 + 8.292 \approx 10.292 \] \[ v_b = \frac{10.292}{1.3} \approx 7.9 , \text{m/s} \]
Step 2: Finding the initial velocity
Now, we can find the initial velocity \( v_0 \) when the ball was thrown. We know that at the bottom of the window (7.5 m), the ball has velocity \( v_b = 7.9 , \text{m/s} \) as it is moving upwards, and it has fallen from some initial height.
Assuming the ball was thrown upwards, we can use the kinematic equation to relate initial velocity \( v_0 \) to \( v_b \): \[ v_b^2 = v_0^2 - 2g(h) \] Where \( h \) is the height difference from the launch point to the bottom of the window (in m).
Rearranging the equation gives: \[ v_0^2 = v_b^2 + 2gh \]
Substituting the values:
- \( v_b = 7.9 , \text{m/s} \)
- \( g = 9.81 , \text{m/s}^2 \)
- \( h = 7.5 , \text{m} \)
Calculating: \[ v_0^2 = (7.9)^2 + 2(9.81)(7.5) \] \[ = 62.41 + 147.15 \approx 209.56 \] \[ v_0 = \sqrt{209.56} \approx 14.47 , \text{m/s} \]
Final answer
Therefore, the initial speed \( v_0 \) at which the ball was thrown is approximately: \[ \boxed{14.5 , \text{m/s}} \]
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