The trajectory is the curved path of the object in respect to its motion along with the gravity. Furthermore, we also refer to it as projectile motion. Besides, we are going to discuss trajectory, trajectory formula, its derivation, and solved examples.

**Trajectory**

It is a form of motion in which an object which is thrown in air travels a curved path under the action of gravity. Also, the projectile is an object which is thrown in the air and the trajectory is the path which it follows. Moreover, the curved path of the object was first revealed by Galileo. Furthermore, the trajectory is a ballistic motion. Besides, gravity is the force of significance that acts on the object.

**Trajectory Formula**

The trajectory formula helps us to find the gravity that acted on an object. Also, the trajectory has vertical (y) and horizontal (x) position components. Moreover, if we launch the projectile with an initial velocity \(v_{0}\), at an angle \(\theta\) from the horizontal plane. Then we can find the vertical position of the object from the horizontal position using this formula

Vertical position = (horizontal position) (tangent of launch angle) – \(\frac{(acceleration due to gravity) (horizontal position)^{2}}{2(initial velocity)^{2}(cosine of launch angle)^{2}}\)

y = x tan \(\theta\) – \(\frac{gx^{2}}{ {2v_{0}}^{2} cos^ {2}\theta}\)

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**Derivation of the Trajectory Formula**

y = refers to the vertical position of the object in meters

x = refers to the horizontal position of the object in meters

\(v_{0}\) = refers to the initial velocity of the object combined with meter per second

g = refers to the acceleration due to gravity that is 9.80 \(m/s^{2}\)

\(\theta\) â€“ refers to the initial angle from the horizontal plane in degrees or radians.

**Solved Examples onÂ Trajectory Formula**

**Example 1**

Firstly, suppose a cricket player hit a ball, guiding it away from the bat at a velocity of 45.0 m/s at an angle of \(66.4^{\circ}\) in relation to the field. Moreover, if the direction of travel of the ball is towards the end of the field which is 140.0 m away. Then what will be the height of the ball when it will reach the end of the field?

**Solution:**

In this question the height of the ball is its vertical position and the horizontal position of attention in this question is end of the field which is x = 140.0 m. Also, we need to solve the vertical position y to know the height of the ball. Moreover, the angle \(\theta\) and the initial velocity (\(v_{0}\)) is present in the question. So, we can use the trajectory formula to solve for y.

y = x tan \(\theta\) – \(\frac{gx^{2}}{{2v_{0}}^{2} cos^{2}\theta}\)

y = (140.0 m) tan (\(66.4^{\circ}\)) – \(\frac{(9.80 m/s^{2})(140.0 m)^{2}}{2(45.0m/s)^{2}cos^{2}(66.4^{\circ})}\)

y \(\cong\) (140.0 m)(2.29)- \(\frac{(9.80 m/s^{2})(140.0 m)^{2}}{2(45.0m/s)^{2}(0.400)^{2}}\)

y \(\cong\) (140.0 m)(2.29)- \(\frac{(9.80 m/s^{2})(19600 m^{2})}{2(2025 m^{2}/s^{2})(0.160)}\)

y \(\cong\) 320.6 m- \(\frac{192080 m^{3}/s^{2}}{648m^{2}/s^{2}}\)

y \(\cong\) 320.6 m- 296.4 m

y \(\cong\) 24.2 m

So, the vertical position of the ball at the end of the field is y \(\cong\) 24.2 m.

**Example 2**

Suppose a water skier plans to set up a stunt in which he is going to jump over a burning obstacle. Also, the fire will be 4.0 m away from the ramp, and it will be 1.0 m taller than the height of the ramp. Moreover, the ramp is inclined at an angle of \(36.9^{\circ}\) in relation to the water. Furthermore, he plans to take off from the ramp at a velocity of 9.0 m/s. So, calculate if he will be able to jump over the flames or not?

**Solution:**

Horizontal position (x) = 4.0 m

Vertical position (y) = ?

Initial velocity )\(v_{0}\)) = 9.0 m/s

Lunch angle = \(36.9^{\circ}\)

Putting values in the trajectory formula

y = x tan \(\theta\) – \(\frac{gx^{2}}{ {2v_{0}}^{2} cos^ {2}\theta}\)

y = (4.0 m) tan (\(36.9^{\circ}\)) – \(\frac{(9.80 m/s^{2})(4.0 m)^{2}}{2(9.0m/s)^{2}cos^{2}(36.9^{\circ})}\)

y \(\cong\) (4.0 m) (0.751)- \(\frac{(9.80 m/s^{2})(4.0 m)^{2}}{2(9.0m/s)^{2}(0.800)^{2}}\)

y \(\cong\) (4.0 m)(0.751)- \(\frac{(9.80 m/s^{2})(16.0 m^{2})}{2(81.0 m^{2}/s^{2})(0.640)}\)

y \(\cong\) 3.0 m – \(\frac{156.8 m^{3}/s^{2}}{103.68 m^{2}/s^{2}}\)

y \(\cong\) 3.0 m- 1.51 m

y \(\cong\) 1.49 m

So, the vertical position of the skier is y \(\cong\) 1.49 m which is more than 1.0 m. Hence, the skier will easily cross the burning obstacle.

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

M=f/g

Interesting studies

It is already correct f= ma by second newton formula…