A frog leaps at [tex]$6.32 m / s$[/tex] at a [tex]$36.3^{\circ}$[/tex] angle to the horizon. What is the maximum height of the frog?
[tex]$\Delta y=[?] m$[/tex]
Answer (1)
Decompose the initial velocity into vertical component: v 0 y = v 0 sin ( θ ) .
Use the kinematic equation v y 2 = v 0 y 2 + 2 a y Δ y and set v y = 0 at maximum height.
Solve for the maximum height: Δ y = 2 g v 0 y 2 = 2 g ( v 0 s i n ( θ ) ) 2 .
Substitute the given values to find the maximum height: Δ y ≈ 0.714 m .
Explanation
Problem Setup We are given the initial velocity v 0 = 6.32 m / s and the launch angle θ = 36. 3 ∘ . We want to find the maximum height Δ y reached by the frog.
Vertical Velocity Component First, we need to decompose the initial velocity into its vertical component v 0 y . We use the formula v 0 y = v 0 sin ( θ ) .
Calculating Vertical Velocity Plugging in the values, we have v 0 y = 6.32 × sin ( 36. 3 ∘ ) .
Kinematic Equation At the maximum height, the vertical velocity v y is zero. We can use the kinematic equation v y 2 = v 0 y 2 + 2 a y Δ y , where a y = − g = − 9.8 m / s 2 is the acceleration due to gravity.
Applying the Equation Substituting v y = 0 and a y = − g into the equation, we get 0 = v 0 y 2 − 2 g Δ y .
Solving for Maximum Height Solving for Δ y , we have Δ y = 2 g v 0 y 2 = 2 g ( v 0 s i n ( θ ) ) 2 .
Substituting Values Plugging in the given values, we have Δ y = 2 ( 9.8 ) ( 6.32 s i n ( 36. 3 ∘ ) ) 2 .
Calculating Maximum Height Calculating the value, we find Δ y ≈ 0.714 m .
Final Answer Therefore, the maximum height of the frog is approximately 0.714 meters.
Examples
Understanding projectile motion, like the frog's leap, is crucial in sports such as basketball or soccer. Calculating the launch angle and initial velocity needed to reach a specific height can optimize a player's performance. Similarly, in engineering, this principle is applied to design trajectories for launching satellites or projectiles, ensuring they reach their intended targets efficiently.
