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Figure 12 shows the path of a projectile launched from ground level with a speed of 25 m s⁻¹ at an angle of 42° to the horizontal - AQA - A-Level Physics - Question 21 - 2017 - Paper 1
Question 21
Figure 12 shows the path of a projectile launched from ground level with a speed of 25 m s⁻¹ at an angle of 42° to the horizontal. What is the horizontal distance f... show full transcript
Worked Solution & Example Answer:Figure 12 shows the path of a projectile launched from ground level with a speed of 25 m s⁻¹ at an angle of 42° to the horizontal - AQA - A-Level Physics - Question 21 - 2017 - Paper 1
Step 1
Calculate the horizontal and vertical components of the velocity
Answer
The horizontal component of the velocity, ( v_{x} ), can be calculated using:
[ v_{x} = v \cdot , \cos(\theta) ]
Substituting the values:
[ v_{x} = 25 , m/s \cdot , \cos(42^{\circ}) \approx 18.66 , m/s ]
The vertical component of the velocity, ( v_{y} ), is calculated as:
[ v_{y} = v \cdot , \sin(\theta) ]
Substituting the values:
[ v_{y} = 25 , m/s \cdot , \sin(42^{\circ}) \approx 16.64 , m/s ]
Step 2
Determine the time of flight
Answer
To find the time of flight, we can use the formula for vertical motion:
[ t = \frac{2 v_{y}}{g} ]
Where ( g ) is the acceleration due to gravity (approximately ( 9.81 , m/s² )). Substituting the values:
[ t = \frac{2 \cdot 16.64}{9.81} \approx 3.39 , s ]
Step 3
Calculate the horizontal distance
Answer
The horizontal distance ( d ) traveled can be calculated using:
[ d = v_{x} \cdot t ]
Substituting the values:
[ d = 18.66 , m/s \cdot 3.39 , s \approx 63.2 , m ]
Therefore, the horizontal distance from the starting point of the projectile when it hits the ground is approximately 63 m.