In this problem, we analyze the motion of a ball thrown towards a rooftop, aiming to determine the height of the roof. The ball is launched at an initial velocity of 13 m/s at an angle of 67.4 degrees. The maximum height of the ball's trajectory occurs directly above the edge of the roof, with the ball landing 3 meters away from this edge.
To solve for the height of the roof, we define the vertical displacement from the launch point (point A) to the point where the ball hits the roof (point C) as Δy from A to C. This can be expressed as the sum of two vertical displacements: Δy from A to B (the maximum height) and Δy from B to C (the downward motion to the roof). Thus, we have:
Δy from A to C = Δy from A to B + Δy from B to C.
To find Δy from A to B, we utilize the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement, which is:
vby² = vay² + 2ayΔyab.
At the peak (point B), the final vertical velocity vby is 0 m/s. The initial vertical velocity vay can be calculated as:
vay = vinitial * sin(67.4°) = 13 * sin(67.4°) ≈ 12 m/s.
Substituting the known values into the kinematic equation gives:
0 = (12)² + 2(-9.8)(Δyab).
Solving for Δy from A to B yields approximately 7.35 meters.
Next, we calculate Δy from B to C. Here, the initial vertical velocity is 0 m/s, and we need to find the time taken for the ball to fall from B to C. We first determine the time using the horizontal motion:
Δxbc = vx * tbc.
Given that the horizontal displacement Δxbc is 3 meters and the horizontal component of the initial velocity vx = vinitial * cos(67.4°) ≈ 5 m/s, we find:
3 = 5 * tbc ⇒ tbc = 0.6 seconds.
Now, we can find Δy from B to C using the equation:
Δybc = vby * tbc + 0.5 * ay * tbc².
Substituting the known values gives:
Δybc = 0 * 0.6 + 0.5 * (-9.8) * (0.6)² ≈ -1.76 meters.
Finally, we combine the two vertical displacements to find the total height of the roof:
Δy from A to C = 7.35 m + (-1.76 m) = 5.59 m.
Thus, the height of the roof is approximately 5.6 meters. This method illustrates how breaking down complex projectile motion problems into manageable intervals can simplify the analysis and lead to accurate results.