A baseball is thrown horizontally from a height of $9.60 \mathrm{~m}$ above the ground with a speed of $25.6 \mathrm{~m} / \mathrm{s}$. Where is the ball after $1.40 \mathrm{~s}$ have elapsed? The ball is above the ground at a horizontal distance of $\mathrm{m}$ from the launch point

Step 1 ：We are given that a baseball is thrown horizontally from a height of 9.60 m above the ground with a speed of 25.6 m/s. We are asked to find where the ball is after 1.40 s have elapsed.

Step 2 ：We first calculate the horizontal distance travelled by the ball using the formula: distance = speed * time. Here, the speed is 25.6 m/s and the time is 1.40 s. So, the horizontal distance = 25.6 m/s * 1.40 s = 35.84 m.

Step 3 ：Next, we calculate the vertical distance travelled by the ball using the formula for the distance travelled by an object under constant acceleration (gravity in this case): distance = initial_velocity * time + 0.5 * acceleration * time^2. Here, the initial vertical velocity is 0 (since the ball is thrown horizontally), the acceleration is -9.8 m/s^2 (the acceleration due to gravity), and the time is 1.40 s. So, the vertical distance = 0 * 1.40 s + 0.5 * -9.8 m/s^2 * (1.40 s)^2 = -9.604 m.

Step 4 ：We then subtract the vertical distance travelled from the initial height to find the height above the ground. Here, the initial height is 9.60 m and the vertical distance is -9.604 m. So, the height above the ground = 9.60 m - (-9.604 m) = 0 m.

Step 5 ：Thus, after 1.40 s, the ball has hit the ground and is approximately 35.84 m away from the launch point.

Step 6 ：Final Answer: The ball is on the ground at a horizontal distance of approximately \(\boxed{35.84}\) meters from the launch point.