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Last updated on September 25, 2025
In physics, the trajectory is the path followed by a projectile under the influence of gravitational forces. The formula to determine the trajectory takes into account the initial velocity, angle of projection, and gravitational acceleration. In this topic, we will explore the formulas related to projectile motion and trajectory.
The trajectory of a projectile is determined by various factors, including initial velocity and angle. Let’s learn the formula to calculate the trajectory of a projectile.
The trajectory of a projectile is given by the equation: \[ y = x \tan(\theta) - \frac{g x^2}{2 (v_0 \cos(\theta))^2} \] where \( y \) is the vertical position, \( x \) is the horizontal position, \( \theta \) is the launch angle, \( g \) is the acceleration due to gravity, and \( v_0 \) is the initial velocity.
The range ( R ) of a projectile is the horizontal distance it covers: [ R = frac{v_0^2 sin(2theta)}{g} ]
The maximum height ( H ) of a projectile is given by: [ H = frac{v_0^2 sin^2(theta)}{2g} ]
In physics and engineering, trajectory formulas are crucial for predicting the path of moving objects.
Here are some important aspects of trajectory formulas:
Students often find physics formulas challenging.
Here are some tips and tricks to master trajectory formulas:
Students make errors when calculating trajectories.
Here are some mistakes and ways to avoid them to master trajectory formulas.
A ball is thrown with an initial velocity of 20 m/s at an angle of 30° above the horizontal. What is the range of the ball?
The range is approximately 34.64 meters.
Using the formula for range: [ R = frac{v_0^2 sin(2theta)}{g} ] [ R =frac{(20)^2 \sin(60°)}{9.8} approx 34.64 , \text{meters} \]
A projectile is launched at an angle of 45° with an initial velocity of 15 m/s. What is the maximum height it reaches?
The maximum height is approximately 5.73 meters.
Using the formula for maximum height: [ H = frac{v_02 \sin2(\theta)}{2g} \] \[ H = \frac{(15)2 \sin2(45°)}{2 × 9.8} \approx 5.73 \, \text{meters} ]
A rock is launched horizontally from a cliff 50 meters high with a speed of 10 m/s. How long does it take to hit the ground?
The time taken is approximately 3.19 seconds.
Using the formula for vertical motion: \[ y = \frac{1}{2} g t^2 \] \[ 50 = \frac{1}{2} × 9.8 × t^2 \] \[ t^2 \approx 10.20 \] \[ t \approx 3.19 \, \text{seconds} \]
A cannonball is fired at an angle of 60° with an initial velocity of 25 m/s. Calculate the time of flight.
The time of flight is approximately 4.41 seconds.
Using the formula for time of flight: [ T = \frac{2v_0 \sin(\theta)}{g} \] \[ T = \frac{2 × 25 × \sin(60°)}{9.8} \approx 4.41 \, \text{seconds} \]
A soccer ball is kicked at an angle of 28° with an initial speed of 18 m/s. How far does it travel horizontally?
The horizontal distance is approximately 29.15 meters.
Using the formula for range: \[ R = \frac{v_02 \sin(2theta)}{g} ] [ R = frac{(18)2 \sin(56°)}{9.8} \approx 29.15 \, \text{meters} \]
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