A baseball is hit and leaves the bat horizontally at a speed of 6.7 m/s. The ball lands 10.3 m from where it was hit. What was the height of the bat at the time it hit the baseball, in m?

t = x/v

y = 1/2gt^2

To find the height of the bat at the time it hit the baseball, we can use the equations of motion for projectile motion along with the known data.

First, we can determine the time it takes for the ball to travel the horizontal distance of 10.3 m. We know that the initial horizontal velocity (Vx) is 6.7 m/s, and the horizontal distance (x) is 10.3 m. We can use the formula:

x = Vx * t

Rearranging the equation, we get:

t = x / Vx

Substituting the given values, we have:

t = 10.3 m / 6.7 m/s

Calculating this, we find that t ≈ 1.537 seconds.

Now, we can find the vertical height (y) of the bat at the time it hit the baseball using the equation of motion for vertical displacement:

y = Vyi * t + 0.5 * g * t^2

Since the ball leaves the bat horizontally, there is no initial vertical velocity (Vyi), so Vyi = 0 m/s. Also, taking the acceleration due to gravity (g) as approximately 9.8 m/s^2, we can substitute the values:

y = 0 m/s * 1.537 s + 0.5 * 9.8 m/s^2 * (1.537 s)^2

Calculating this, we find that y ≈ 11.54 meters.

Therefore, the height of the bat at the time it hit the baseball is approximately 11.54 meters.