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# Equation of trajectory , how to find what is the equation of trajectory of a projectile

By   April 10, 2023

find Equation of trajectory , how to find what is the equation of trajectory of a projectile ?

Motion in Two Dimensions : Projectile Motion

(i) A body projected horizontally with a velocity u from a height h. [Fig. below] Horizontal and vertical distances covered in time t are

Differentiating Eqs. (2.5) and (2.6) w.r.t time t, we get the horizontal and vertical velocities.

Equation to trajectory

Eliminating t from Eqs. (2.5) and (2.6), we get

Since y ∝ x2 , the trajectory of the body is parabolic.

Time of flight (tf )

Putting y = h and t = tf in Eq. (2.6), we get tf

Horizontal range (R)
Putting t = tf and x = R in Eq. (2.5), we get

Magnitude of resultant velocity at time t is

The angle α which the resultant velocity vector subtends with the vertical is given by

(ii) A body projected from the ground with a velocity u at an angle q with the horizontal. (Fig. 2.21) The horizontal and vertical distances covered in time t are

Horizontal and vertical components of the velocity at time t are

Magnitude of resultant velocity at time t is

The angle α subtended by the resultant velocity vector with the horizontal is given by

Equation of trajectory

Eliminating t from Eqs. (2.8) and (2.9), we get

Time of flight (tf)
Putting y = 0 and t = tfin Eq. (2.9), we get

Maximum height attained (hmax)
Put t = tf / 2 and y = hmax in Eq. (2.9), we get

Horizontal range (R)
Putting t = tf and x = R in Eq. (2.8) we get

(iii) A body projected from a height h with a velocity u at an angle θ with the horizontal.

If T is the total time of flight, then we have

The horizontal range is R’ = (u cos θ)T

Applications

(i) The horizontal range is the same for angles θ and (90° – θ).

(ii) The horizontal range is maximum for θ = 45°

Rmax = u2/g

(iii) When horizontal range is maximum, hmax = Rmax / 4

(vi) To find R and hmax from the equation of trajectory

y = ax – bx2

(a) At O and B, y = 0. Putting y = 0 in the above equation, we have 0 = ax – bx

x = 0

x = a/b Therefore R = a/b

(vii) If A and B are two points at the same horizontal level on a trajectory at a height h from the ground, (see Fig. 2.24), then

(c) Average velocity during time interval (t2 – t1) is v = u cos θ (∵ during this interval, the vertical displacement is zero)

(viii) Velocity and Direction of Motion of Projectile at any Height. Let P be a point on the trajectory of a projectile at a height h and let v be the velocity of the projectile at that height. If α is the angle which the velocity vector makes with the horizontal, then the horizontal and vertical components of the velocity are given by vx = ux = constant

This gives the speed of the projectile at height h. The direction of the velocity vector (i.e., direction of motion) is obtained by taking the square root of Eq. (ii) and then dividing by Eq. (i). We get

(ix) Time of Flight and Range of a Projectile on an Inclined Plane Consider an inclined plane OAB making an angle a with the horizontal (Fig. 2.25). Let a body be projected with a velocity u at an angle q with the horizontal. Let us choose the x-axis along the plane OA and y-axis perpendicular to the plane OA. Let the body hit the inclined plane at point P so that R = OP is the range on the inclined plane. The x and y components of the velocity of the projectile are

The x and y components of acceleration due to gravity are – g sin a and – g cos a respectively, as shown in Fig. 2.25. Let Tf be the time of flight on the inclined plane. Since the net vertical displacement in time Tf is zero (i.e., h = 0), we have

During this time, the horizontal component of velocity u cos θ remains constant. Hence, horizontal distance OQ is

Range of the projectile on the inclined plane is

Relative Velocity in Two Dimensions

The relative velocity of a body B with respect to body A is defined as

If vectors vA and vB are inclined to each other at an angle θ as shown in Fig. 2.31, the relative velocity vBA is found as follows

Thus, the magnitude and direction of vector vBA can be found by finding the resultant of vectors vB and –vA which is vector OC as shown in Fig. 2.31. Magnitude of vector vBA is given by

The angle b which the resultant vector OC subtends with vector OD is given by

Special Case

(i) If vector vA and vB are in the same direction, θ = 0°, then

(ii) If vector vA and vB are in opposite direction, θ = 180°, then vBA = vB + vA.

Applications

(i) To cross the river of width d along the shortest path which is PQ, the boat must move along PR making an angle (90° + θ) with the direction of the stream such that the direction of the resultant velocity v is along PQ. Angle θ is given by (see Fig. 2.32)

The time taken to cross the river along the shortest path is given by

(b) To cross the river in the shortest time, the boat should move along PQ. The shortest time is given by

At this time, the boat will reach the point R on the opposite bank of the river at a distance x from the point Q (Fig. 2.33). From the Figure, we have

x = d tan θ

(ii) Holding an Umbrella to Project from Rain Let vr be the velocity of the rain falling vertically downward and vm the velocity of a man walking from north

to south direction (Fig. 2.34). In order to protect himself from rain, he must hold his umbrella in the direction of the resultant velocity v, which is given by

This is the speed with which the rain strikes the umbrella. If θ is the angle subtended by the resultant velocity v with the vertical, then from triangle ORM¢, we have

Thus, the man must hold the umbrella at an angle θ with the vertical towards north.

Uniform Circular Motion

(i) For a body moving in a horizontal circle The centripetal acceleration of a body of mass m moving in a circle of radius R with a constant speed v (or angular speed ω) is

(ii) For a body moving in a vertical circle The minimum speed to complete the circle when the body is at the top of the circle is v = √Rg . The minimum speed to complete the circle when the body is at the bottom of the circle is v = √5Rg .

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