A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. The time that elapses from when the ball passes the bottom of the window on the way up, to when it passes the top of the window on the way back down is 1.3 s.
What was the ball’s initial speed, in meters per second?
To find the ball's initial speed, we can use the following kinematic equation:
\(y = y_0 + v_0t + \frac{1}{2}gt^2\)
where:
y = final position (0 meters)
y_0 = initial position (7.5 meters)
v_0 = initial velocity (what we're trying to find)
t = time (1.3 seconds)
g = acceleration due to gravity (-9.8 m/s^2)
Since the ball reaches a maximum height at the top of the window, the final position is 2 meters above the initial position. Plugging in the known values, we get:
\(0 = 7.5 + v_0(1.3) + \frac{1}{2}(-9.8)(1.3)^2\)
Simplifying the equation, we have:
\(0 = 7.5 + 1.3v_0 - 8.053\)
Rearranging the equation to solve for \(v_0\), we have:
\(1.3v_0 = 0 - 7.5 + 8.053\)
\(1.3v_0 = 0.553\)
\(v_0 = \frac{0.553}{1.3}\)
\(v_0 \approx 0.425\) m/s
Therefore, the ball's initial speed was approximately 0.425 meters per second.