Examples of Projectiles Simplified Revision Notes for Scottish Highers Physics
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Learn about Projectiles for your Scottish Highers Physics Exam. This Revision Note includes a summary of Projectiles for easy recall in your Physics exam
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Examples of Projectiles
Examples of Projectiles
Introduction
When objects are launched into the air with a certain velocity and angle, their motion can be described as projectile motion.
Understanding key parameters like time of flight, maximum height, and range is essential for analysing such motion.
Given Information
A girl kicks a ball into the sea at a velocity (v) of 7.2 m/s at an angle (θ) of 30° above the horizontal.
Vertical and Horizontal Components
To analyse the motion, we must calculate the vertical and horizontal components of the initial velocity.
The vertical component (v_vertical_initial) is calculated as:
v_vertical_initial = v × sin(θ)
The horizontal component (v_horizontal_initial) is calculated as:
v_horizontal_initial = v × cos(θ)
Time of Flight
The time of flight (t_flight) can be determined using equations of motion since the vertical acceleration is uniform.
Vertical velocity (v_vertical) at the highest point is 0 m/s, so we can use the equation:
Since v_vertical_initial is positive and a_vertical (acceleration due to gravity) is negative, we get:
0 = v_vertical_initial - g × t_flight
Solving for t_flight, we find:
t_flight = (v_vertical_initial)/g
The total time of flight is then:
t_flight_total = 2 × t_flight (as motion is symmetrical)
Examples of Projectiles
Maximum Height
To calculate the maximum height (h_max), we need to know the time taken to reach it.
At the maximum height, the vertical velocity is 0 m/s, so we can use the equation:
v_vertical = v_vertical_initial - g × t_to_max
Solving for t_to_max, we find:
t_to_max = (v_vertical_initial)/g
The maximum height reached is then:
h_max = (v_vertical_initial²)/(2 × g)
Range
Finally, the range (R) is calculated using the horizontal component (v_horizontal_initial) and the total time of flight (t_flight_total):
R = v_horizontal_initial × t_flight_total
Calculations
Given v = 7.2 m/s and θ = 30°, we calculate:
v_vertical_initial = 7.2 m/s × sin(30°) = 3.6 m/s
v_horizontal_initial = 7.2 m/s × cos(30°) = 6.24 m/s
Using g = 9.8 m/s²:
t_flight = (3.6 m/s)/(9.8 m/s²) ≈ 0.367 s
t_flight_total = 2 × 0.367 s = 0.734 s
Calculating h_max:
h_max = (3.6 m/s)²/(2 × 9.8 m/s²) ≈ 0.656 m
Calculating R:
R = 6.24 m/s × 0.734 s ≈ 4.57 m
bookmarkSummary
In Summary
When analysing projectile motion, parameters such as time of flight, maximum height, and range can be calculated.
These calculations involve understanding the vertical and horizontal components of motion, uniform acceleration due to gravity, and equations of motion.
In the given example, the time of flight is approximately 0.734 seconds, the maximum height is about 0.656 meters, and the range is approximately 4.57 meters.
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