Let's solve the problem step-by-step: --- **Given:** - Time, \( t = 4.70 \) s - Height, \( h = 22.0 \) m above release level - Angle just before landing with roof, \( \theta = 59.0^\circ \) - Acceleration due to gravity, \( g = 9.8 \) m/s² ### (a) **Find the horizontal distance \( d \) it travels.** #### **Step 1: Analyzing the final velocity** Let’s denote the final velocity components just before the ball lands as \( v_{x,f} \) (horizontal) and \( v_{y,f} \) (vertical). Since the angle given is with the roof (horizontal): \[ \tan\theta = \frac{v_{y,f}}{v_{x,f}} \] \[ \tan(59.0^\circ) = \frac{v_{y,f}}{v_{x,f}} \] \[ v_{y,f} = v_{x,f} \cdot \tan(59.0^\circ) \] #### **Step 2: Vertical motion analysis** Let’s take down as positive (since we’re "playing the tape backward", as the hint says). Over the time \( t \), the ball moves from height \( h \) to ground (a displacement of \( -h \), if up is positive): \[ y_f = y_0 + v_{y,0} t - \frac{1}{2}gt^2 \] But since we are considering the landing point as the final point, and the release point as the initial: \[ h = v_{y,i} t - \frac{1}{2} g t^2 \] So, \[ v_{y,i} = \frac{h}{t} + \frac{1}{2}g t \] But we need the velocity **just before landing**. The vertical velocity at landing: \[ v_{y,f} = v_{y,i} - g t \] But let's reverse the motion, as per hint. So, the velocity vector just before landing (when the ball lands on the roof) will be the same as the initial velocity if we run the motion backward. #### **Step 3: Find \( v_{y,f} \) and \( v_{x,f} \) in terms of time** From vertical displacement: \[ h = v_{y,f} t + \frac{1}{2} g t^2 \] But since on the roof, the velocity is directed downward at angle \( \theta \), we can say \[ v_{y,f} = v_{x,f} \tan(59.0^\circ) \] #### **Step 4: Express everything in terms of \( v_{x,f} \) and solve for \( v_{x,f} \):** \[ h = v_{y,f} t + \frac{1}{2} g t^2 \] \[ h = (v_{x,f} \tan(59.0^\circ)) t + \frac{1}{2} g t^2 \] \[ v_{x,f} = \frac{h - \frac{1}{2} g t^2}{t \tan(59.0^\circ)} \] Now plug in the values: - \( h = 22.0 \) m - \( g = 9.8 \) m/s² - \( t = 4.70 \) s - \( \tan(59.0^\circ) \approx 1.664 \) \[ v_{x,f} = \frac{22.0 - 0.5 \cdot 9.8 \cdot (4.70)^2}{4.70 \cdot 1.664} \] Calculate \( 0.5 \cdot 9.8 \cdot (4.70)^2 \): \[ (4.70)^2 = 22.09 \] \[ 0.5 \cdot 9.8 \cdot 22.09 = 4.9 \cdot 22.09 = 108.24 \] So, \[ v_{x,f} = \frac{22.0 - 108.24}{4.70 \cdot 1.664} \] \[ v_{x,f} = \frac{-86.24}{7.8208} \] \[ v_{x,f} = -11.03 \text{ m/s} \] The negative sign indicates the direction but let's use the magnitude: \[ v_{x,f} = 11.03 \text{ m/s} \] #### **Step 5: Find horizontal distance \( d \):** \[ d = v_{x,f} \cdot t \] \[ d = 11.03 \times 4.70 = 51.84 \text{ m} \] --- ### (b) **Magnitude of the ball’s initial velocity** The initial velocity (reversed, i.e., final velocity at the roof): \[ v_f = \sqrt{(v_{x,f})^2 + (v_{y,f})^2} \] Recall: \[ v_{y,f} = v_{x,f} \tan(59^\circ) = 11.03 \times 1.664 = 18.35 \text{ m/s} \] \[ v_f = \sqrt{(11.03)^2 + (18.35)^2} \] \[ v_f = \sqrt{121.66 + 336.72} \] \[ v_f = \sqrt{458.38} = 21.42 \text{ m/s} \] --- ### (c) **Angle of the ball’s initial velocity (relative to the horizontal)** \[ \theta_i = \arctan\left(\frac{v_{y,f}}{v_{x,f}}\right) \] \[ \theta_i = \arctan\left(\frac{18.35}{11.03}\right) \] \[ \theta_i = \arctan(1.664) = 59.0^\circ \] --- ## **Final Answers:** ### (a) **Horizontal distance \( d \):** \[ \boxed{51.8 \ \text{m}} \] ### (b) **Magnitude of the ball’s initial velocity:** \[ \boxed{21.4 \ \text{m/s}} \] ### (c) **Angle of the ball’s initial velocity (relative to the horizontal):** \[ \boxed{59.0^\circ} \] --- **Let me know if you need clarifications or further explanations!**

The object moves under gravity and reaches a certain height after a specific time, with its velocity components related by the given angle. Its horizontal and vertical motions can be analyzed by reversing the motion, considering the initial and final velocities. The initial launch speed and angle can be found using the velocity components derived from the motion equations.