To determine the time it takes for the baseball to hit the ground, we can break down the initial velocity into horizontal and vertical components.
The horizontal component of the velocity (v_x) remains constant throughout the motion because there are no horizontal forces acting on the baseball. The horizontal component can be found using the formula:
v_x = v * cos(theta)
where v is the magnitude of the initial velocity (32.0 m/s) and theta is the launch angle (25.0 degrees).
v_x = 32.0 m/s * cos(25.0 degrees) v_x ≈ 28.876 m/s
The vertical component of the velocity (v_y) changes due to the acceleration of gravity. We can find the time it takes for the baseball to reach the ground by using the formula:
y = v_y * t + (1/2) * g * t^2
where y is the vertical displacement (115 m), v_y is the vertical component of the velocity, g is the acceleration due to gravity (-9.8 m/s²), and t is the time.
At the highest point of the trajectory, the vertical velocity becomes zero, so we can find v_y using:
v_y = v * sin(theta)
v_y = 32.0 m/s * sin(25.0 degrees) v_y ≈ 13.630 m/s
Substituting the known values into the equation for vertical displacement:
115 m = (13.630 m/s) * t + (1/2) * (-9.8 m/s²) * t^2
Simplifying the equation:
-4.9 t^2 + 13.630 t - 115 = 0
To solve for t, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = -4.9, b = 13.630, and c = -115.
Plugging in these values:
t = (-13.630 ± √((13.630)^2 - 4 * -4.9 * -115)) / (2 * -4.9)
After calculating the expression inside the square root and performing the calculations, we find two solutions for t:
t ≈ 4.04 seconds (ignoring the negative solution)
Therefore, it takes approximately 4.04 seconds for the baseball to hit the ground.