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A projectile is launched with a speed of 25 m s⁻¹ at an angle of 35° to the horizontal, as shown in the diagram - AQA - A-Level Physics - Question 23 - 2018 - Paper 1

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A projectile is launched with a speed of 25 m s⁻¹ at an angle of 35° to the horizontal, as shown in the diagram. Air resistance is negligible. What is the time tak... show full transcript

Worked Solution & Example Answer:A projectile is launched with a speed of 25 m s⁻¹ at an angle of 35° to the horizontal, as shown in the diagram - AQA - A-Level Physics - Question 23 - 2018 - Paper 1

Step 1

Step 1: Calculate the vertical component of the velocity

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Answer

To find the vertical component of the initial velocity, we can use the formula:

vy=vimesextsin(heta)v_{y} = v imes ext{sin}( heta)

where:

  • v=25 extms−1v = 25 \, ext{m s}^{-1} (initial speed)
  • θ=35°\theta = 35° (angle of launch)

Therefore, vy=25imesextsin(35°)≈25imes0.5736≈14.34 extms−1v_{y} = 25 imes ext{sin}(35°) \approx 25 imes 0.5736 \approx 14.34 \, ext{m s}^{-1}

Step 2

Step 2: Determine the time taken to reach the highest point

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Answer

At the highest point, the vertical velocity becomes zero. We can use the equation:

t=vygt = \frac{v_{y}}{g}

where:

  • g=9.81 extms−2g = 9.81 \, ext{m s}^{-2} (acceleration due to gravity)

Thus, t=14.349.81≈1.46 extst = \frac{14.34}{9.81} \approx 1.46 \, ext{s}

Step 3

Step 3: Total time for the projectile to return to the ground

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Answer

The total time for the projectile to return to the ground is twice the time taken to reach the highest point:

T=2t≈2×1.46≈2.92 extsT = 2t \approx 2 \times 1.46 \approx 2.92 \, ext{s}

Rounding to one decimal place, this gives approximately 2.9 s.

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