a football is kicked at an velocity at an angle of 37.0° with an unknown velocity and lands 45.0 meters down field.a. find the maximum height of the footballb. the time to reach maximum heightc. the time in air before striking the ground
Answer (1)
Answer:a) Max height = 8.48 mb) Time to max height = 1.32 sc) Total time in air = 2.64 s______________________________________Given Data:Angle: θ = 37.0° Range: R = 45.0 mGravity: g = 9.81 m/s²1: Find Initial Velocity (v₀)[tex] \tt{v_0^2 = \frac{R \cdot g}{\sin(2\theta)}}[/tex][tex] \tt{v_0^2 = \frac{45.0 \times 9.81}{\sin(74.0°)}}[/tex][tex] \tt{v_0^2 = \frac{441.45}{0.9613}}[/tex][tex] \tt{v_0 = \sqrt{459.3} \approx 21.45 \: { m/s}}[/tex]2: Find Maximum Height (hₘₐₓ)[tex] \tt{v_{0y} = v_0 \sin\theta = 21.45 \times 0.6018 = 12.9 \text{ m/s}}[/tex][tex] \tt{h_{\ {max}} = \frac{(v_{0y})^2}{2g} = \frac{(12.9)^2}{2 \times 9.81}}[/tex][tex] \tt{h_{\ {max}} = \frac{166.41}{19.62} \approx 8.48 \: { m}}[/tex]3: Find Time to Max Height (tₘₐₓ)[tex] \tt{t_{\ {up}} = \frac{v_{0y}}{g} = \frac{12.9}{9.81} \approx 1.32 \ { s}}[/tex]4: Find Total Time in Air (tₜₒₜₐₗ) [tex] \tt{t_{\ {total}} = 2 \times t_{\ {up}} = 2 \times 1.32 = 2.64 \ { s}}[/tex]
