POSITION AND SPEED MID FLIGHT

Decompose the launch: vₓ = 45·cos(37°) ≈ 35.93 m/s, vy₀ = 45·sin(37°) ≈ 27.08 m/s. The total time of flight is T = 2·vy₀/g = 2·27.08/9.807 ≈ 5.52 s. At t_mid = T/2 ≈ 2.76 s the horizontal position is x_mid = vₓ·t_mid ≈ 35.93·2.76 ≈ 99.2 m. The vertical position is y_mid = vy₀·t_mid - ½·g·t_mid² ≈ 27.08·2.76 - ½·9.807·2.76² ≈ 74.74 - 37.36 ≈ 37.4 m. The vertical velocity at this moment is vy(t_mid) = vy₀ - g·t_mid = 27.08 - 9.807·2.76 ≈ 0. It is exactly zero — the ball has reached its peak. This is not a coincidence: the ball rises for exactly half the flight time and falls for the other half, by the symmetry of the parabola. At the peak the velocity is entirely horizontal: speed_mid = vₓ ≈ 35.93 m/s. This is also the minimum speed during the flight — the ball is moving slowest at the top. After the peak, vy rebuilds in the downward direction, so the speed increases again until landing. The y-coordinate at mid-flight equals the peak height: H = vy₀²/(2g) ≈ 27.08²/19.614 ≈ 37.4 m, confirming that mid-flight in time and maximum height are the same event.