A projectile is fired at an angle with horizontal θ. Show that its trajectory is a parabola.

Long Question Solution

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A projectile is fired at an angle with horizontal θ. Show that its trajectory is a parabola.

Kinematics - Long Question Answer 1

Long Question Answer

Question: A projectile is fired at an angle with horizontal θ. Show that its trajectory is a parabola. Also derive expressions for the maximum height attained, the time of flight, and horizontal range.

Answer: Let us consider a projectile having mass 'm' is projected with a velocity ‘u’ at an angle θ with horizontal. The projectile goes up with decreasing velocity, reaches the highest point, and falls back to the level of projection as shown in the figure.

trajectory

If the air resistance is negligible, the horizontal component u, remains constant, throughout the motion whereas vertical component goes on decreasing and becomes zero at the highest point of its path then finally it increases downwards due to gravitational attraction. 

The horizontal distance travelled by projectile t instant time ‘t’ is

S = ux.t [s = u.t + ½ at2 and ax = 0]

x = ucosθ.t   ……………………. (I) 

where x is the distance travelled by a projectile in instant time ‘t’ and ucosθ is the horizontal velocity of the projectile at an instant time ‘t’.

The vertical distance travelled by the projectile at an instant time ‘t’ is 

S = u.t – 1/2gt2  ……………… (II)

From equation (I),  t = x/ucosθ

Y = usinθ.x/ucosθ – ½. g. (x/ucosθ)2

Y = x.tanθ – (1/2u2. g/cos2θ)

Y = Ax + Bx2  ………………………. (III)

Where, A = tanθ and B = -1/2 g/cos2θ. u2

Equation (III) represents equation of parabola. Hence, trajectory of a projectile is a parabola.

a. Maximum height (hmax): The maximum vertical distance covered by a projectile during its flight is called maximum height. The vertical component of velocity vy is zero in maximum height. (i.e. vy = 0)

We have,

V2 = u2 + 2as

Or, 0 = (usinθ)2 + 2. (-g) .h

Or, 2gh = u2sin2θ

Or, h = u2sin2θ/2g  ……………... (iv)

b. Time of flight (T): It is a total time for which the projected projectile remains in the air. At the end of flight projectile reaches to the ground, so vertical distance traveled by the projectile is zero i.e. h = 0.

0 = usinθ. T – ½. gT2

½. gT2 = usinθ. T

T = 2usinθ/g  ………(V)

c. Horizontal Range(R): The horizontal distance covered by the projectile body during its time of flight is called Horizontal Range. It is equal to the product of a horizontal component of velocity and the total time of flight.

R = ux.T

R = ucosθ. 2usinθ/g

R = u2 sin2θ/g ……………. (VI)

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