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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

Question 23

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

Calculate the vertical component of the initial velocity

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Answer

To find the time taken for the projectile to return to the ground, we first need to calculate the vertical component of the initial velocity ( v_{y} ). This can be found using the formula:

$v_{y} = v \sin(\theta)$

Where ( v = 25 \text{ m s}^{-1} ) and ( \theta = 35° ).

Substituting the values: $v_{y} = 25 \sin(35°) \approx 25 \times 0.5736 \approx 14.34 \text{ m s}^{-1}$

Step 2

Calculate the total time of flight

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Answer

The time of flight for a projectile can be calculated with the formula:

$t = \frac{2 v_{y}}{g}$

Where ( g \approx 9.81 , \text{m s}^{-2} ) is the acceleration due to gravity. Substituting the value of ( v_{y} ): $t = \frac{2 \times 14.34}{9.81} \approx \frac{28.68}{9.81} \approx 2.92 \, \text{s}$

Rounding to the nearest tenth gives us approximately 2.9 s.

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