A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.20 s for the ball to reach its maximum height.
To find out how high the baseball travels, we first need to determine its initial velocity when it was hit by the bat. To do this, we can use the fact that the time taken for the ball to reach its maximum height is equal to half of the total time it takes to reach the ground (since the upward and downward flights take the same amount of time).
Let's call the total time taken for the ball to reach its maximum height and return to the ground as T.
We know that the time taken to reach the maximum height is 3.20 s, so we can write:
T = 2 × 3.20 s
T = 6.40 s
Now, we can calculate the time taken for the ball to travel upwards to its maximum height. Since it takes half of the total time, we have:
t_up = 6.40 s / 2
t_up = 3.20 s
Next, we need to calculate the initial velocity of the ball. To do this, we can use the equation of motion:
h = v_i * t + (1/2) * a * t^2
Since the ball is at its maximum height, the final height (h) is zero. Also, the acceleration (a) due to gravity is -9.8 m/s^2 (assuming no air resistance).
0 = v_i * t_up + (1/2) * (-9.8 m/s^2) * (t_up)^2
Simplifying the equation gives us:
0 = v_i * 3.20 s - 4.9 m/s^2 * (3.20 s)^2
0 = v_i * 3.20 s - 4.9 m/s^2 * 10.24 s^2
0 = v_i * 3.20 - 50.176 m
Now, let's solve for the initial velocity (v_i):
v_i = 50.176 m / 3.20 s
v_i ≈ 15.67 m/s
Therefore, the initial velocity of the baseball when it was hit by the bat is approximately 15.67 m/s.