A baseball is hit with a velocity of 18.56 m/s of at an angle of 39.7° to the horizontal and returns to the same initial height.Calculate the time of flight in seconds.Give your answer to 2 decimal places without units.

Question

BlackTom AI · Accepted Answer

Time of flight for a projectile returning to the same height is T = 2u_y/g, where u_y is the initial vertical component of velocity and g is the acceleration due to gravity.

Relevant knowledge points:
- Decompose the initial velocity into vertical and horizontal components: u_y = u sinθ.
- For symmetric projectile motion (same launch and landing height) and neglecting air resistance, total flight time T = 2u_y/g.
- Use g = 9.81 m/s^2.

Step-by-step calculation:
1. Given u = 18.56 m/s and θ = 39.7°, compute the vertical component:
   u_y = 18.56 × sin(39.7°).
   Compute sin(39.7°) ≈ 0.638769 (using a calculator or trig table).
   So u_y ≈ 18.56 × 0.638769 = 11.8556 m/s (intermediate value).

2. Apply the time-of-flight formula:
   T = 2 u_y / g
     = 2 × 11.8556 / 9.81
     = 23.7112 / 9.81
     ≈ 2.4179 s.

3. Round to two decimal places (as requested): T ≈ 2.42.

Why this is correct:
- The vertical motion satisfies y(t) = u_y t − (1/2) g t^2. Setting y = 0 (return to initial height) gives t( u_y − (1/2) g t ) = 0, so the nonzero root is t = 2 u_y / g. This derivation justifies the formula used.
- Numerical substitution and arithmetic lead to 2.4179 s, which rounds to 2.42 s.

Other options: Not applicable (this is a numerical response, not multiple choice).

Conclusion: The calculated time of flight is 2.42, which matches the provided correct answer.

A baseball is hit with a velocity of 18.56 m/s of at an angle of 39.7° to the horizontal and returns to the same initial height.Calculate the time of flight in seconds.Give your answer to 2 decimal places without units.数值题

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