## Step 1: Find the Time of Flight Given: - Initial speed, \( v_ = 44 \) m/s - Launch angle, \( \theta = 37^\circ \) - Horizontal distance, \( X = 48 \) m The horizontal velocity is: \[ v_{x} = v_ \cos \theta = 44 \cos(37^\circ) \approx 44 \times .7986 \approx 35.1 \text{ m/s} \] Time to cover \( X \): \[ X = v_{x} t \implies t = \frac{X}{v_{x}} = \frac{48}{35.1} \approx 1.37 \text{ s} \] **Explanation:** We found the time it takes for the projectile to land by using horizontal motion (distance = speed × time). --- ## Step 2: Find the Height of the Landing Point Vertical motion equation: \[ y = v_{y} t - \frac{1}{2} g t^2 \] where \( v_{y} = v_ \sin \theta = 44 \sin(37^\circ) \approx 44 \times .6018 \approx 26.5 \) m/s. Plug values in: \[ y = (26.5 \times 1.37) - \frac{1}{2} (9.81) (1.37)^2 \] \[ y = 36.3 - 9.2 = 27.1 \text{ m} \] **Explanation:** The vertical position at time \( t \) gives the height of the landing point above the launch point. --- ## **Final Answer** \[ \boxed{27.1 \text{ m}} \] The height of the landing point is **27.1 meters**.

# Projectile Motion Problem ## Step 1: Find the Time of Flight ### Given Data: - Initial Speed, \( v_ = 44 \) m/s - Launch Angle, \( \theta = 37^\circ \) - Horizontal Distance, \( X = 48 \) m ### Calculation of Horizontal Velocity: The horizontal component of the initial velocity is calculated using: \[ v_{x} = v_ \cos \theta = 44 \cos(37^\circ) \approx 44 \times 0.7986 \approx 35.1 \text{ m/s} \] ### Time to Cover \( X \): Using the formula for horizontal distance: \[ X = v_{x} t \implies t = \frac{X}{v_{x}} = \frac{48}{35.1} \approx 1.37 \text{ s} \] **Explanation:** We determined the time it takes for the projectile to land by analyzing its horizontal motion, where distance equals speed multiplied by time. --- ## Step 2: Find the Height of the Landing Point ### Vertical Motion Equation: The vertical motion can be described by: \[ y = v_{y} t - \frac{1}{2} g t^2 \] where the vertical component of the initial velocity is: \[ v_{y} = v_ \sin \theta = 44 \sin(37^\circ) \approx 44 \times 0.6018 \approx 26.5 \text{ m/s} \] ### Plugging Values into the Equation: Substituting the values into the vertical motion equation: \[ y = (26.5 \times 1.37) - \frac{1}{2} (9.81) (1.37)^2 \] Calculating each term: \[ y = 36.3 - 9.2 = 27.1 \text{ m} \] **Explanation:** The vertical position at time \( t \) gives us the height of the landing point above the launch point. --- ## **Final Answer** \[ \boxed{27.1 \text{ m}} \] The height of the landing point is **27.1 meters**.