a softball batter hits a fly ball that is caught 3.8 seconds later, 88m from home plate. The ball is caught at the same height from which it was thrown.
What was its initial velocity?
What is the maximum height above its launch point that it reached?
the velocity is the vector sum of the horizontal and vertical components
horizontal ... 88 m / 3.8 s
vertical ... time up equals time down
... (3.8 s / 2) * 9.8 m/s^2
the max height is half the initial vertical component, multiplied by half the flight time
To determine the initial velocity of the softball and the maximum height it reached, we can use the principles of projectile motion. In this case, we can assume that the softball was launched horizontally, as there is no mention of an initial vertical velocity.
1. Initial velocity (Vi):
Since the ball was caught 3.8 seconds after being hit, we can use the formula for horizontal distance: distance = velocity × time. In this case, the horizontal distance is 88m, and the time is 3.8 seconds. Since the ball was launched horizontally, the initial vertical velocity is zero.
Distance = Velocity × Time
88m = Vix × 3.8s
The initial velocity in the horizontal direction (Vix) can be calculated by rearranging the formula:
Vix = Distance / Time
Vix = 88m / 3.8s
Calculating Vix:
Vix ≈ 23.16 m/s
So, the initial velocity of the softball is approximately 23.16 m/s.
2. Maximum height (Hmax):
To find the maximum height, we need to calculate the time it takes for the ball to reach its highest point. Since there is no initial vertical velocity, the time taken to reach the highest point will be half of the total time of flight (Tf).
Tf = 2 × time
Tf = 2 × 3.8s
Calculating Tf:
Tf = 7.6 s
Now we can use the formula for vertical displacement at the highest point, assuming that the acceleration due to gravity is -9.8 m/s² (taking downward as the negative direction):
Vertical displacement (Δy) = (Initial vertical velocity × time) + (0.5 × acceleration × time²)
0 = (0 × t) + (0.5 × -9.8 × t²)
Simplifying the equation:
0 = -4.9t²
The equation simplifies to a parabolic equation with one root. Solving for t², we have:
t² = 0
Thus, t = 0. Since we are using the quadratic equation, we take the positive root. Therefore, t = 0.
Since t = 0, the ball reaches the highest point immediately after being hit. Therefore, the maximum height above its launch point is zero.
In summary:
- The initial velocity (Vix) of the softball is approximately 23.16 m/s.
- The maximum height above its launch point is zero.