Projectile Motion

1. Projectile Motion : 

 For vertical displacement,

y = usinθ t - ½gt²……………… (i)

For horizontal displacement,

X = ucosθ t

t = x/ucosθ …………………. (ii)

From equation (i) and (ii)

y = usinθ.x/ucosθ – ½g(x/ucosθ).

y = x tanθ - g/(2u²cos²θ)

is similar to y = ax + bx² so the path of projectile is parabolic.

2. Time of flight (T) :

T = 2usinθ/g

Time of ascent and time of descent are same i.e. t = usinθ/g.

3. Maximum Height (H) :

H = u²sin²θ/2g.

4. Range of Projectile (R) :

R = u²sin2θ/g.

R_max = u²/g.

5. Maximum height for the projectile with maximum range :

When a projectile is fired at an angle of 45° then,

R_max = u²/g.

For maximum height

R_max = 4H

6. Two Angle of projection for same horizontal range :

A projectile is fired from ground with velocity u at an angle of θ = with horizontal then

Range (R₁) = u²sin2θ/g

For another angle of projection for same range with same velocity will be

Horizontal range (R₁) = u²sin2θ/g

= u²/g.sin(180°- 2θ)

= u²/g.sin{2(90°- θ)}

= u²/g.sin2θ= R₂

θ and (90°- θ) are the two angle of projection for a projectile with same range with same velocity.

7. Velocity and direction of projectile at any height :

The horizontal component of projectile remain constant through the motion but vertical component is accelerating. At any point P, at horizontal displacement y

Horizontal velocity (V_x) = ucosθ

Vertical velocity (V_y) is given by

V_y²= (usinθ)²- 2gy

or, V_y²= √(u²sin²θ- 2gy)

Resultant velocity (V) = √(V_x² + V_y²)

= √(u² – 2gy)

∴V = √(u² – 2gy)

For direction α be the direction of resultant with horizontal,

So,

tanθ = v_y/v_x = {√(u²sin²θ-2gy)}/{ucosθ}

KE of projectile of mass m is, KE = ½mv² = ½m(V_x²+ V_y²).

Some important tips :

- In projectile motion acceleration is due to gravity.

- The path of the projectile is called trajectory.

- Nature of trajectory is parabolic.

- Horizontal component of velocity is constant through out the motion.

- If a projectile is projected so that its range obtained is maximum.then maximum height attained by it is ^th of maximum range.

- If a person can throw maximum horizontal distance R₀, then he can throw maximum height R₀/2.

- Height is maximum if θ = 90°, H_max = R_max.

- At heighest point angle between velocity and acceleration is 90°.

- Velocity is minimum at highest point hence kinetic energy is minimum.

- If a projectile is thrown with speed u at angle θ with horizontal the projectile makes an angle ‘α’ with horizontal then its speed, v = ucosθ.Secα.

- Average velocity during time of ascent (i.e. average velocity when projectile is at highest point).

V_avg = u/2.√( 1 + 3cos²θ)

- Average velocity when projectile strikes to ground,

V_avg.= ucosθ

- Change in speed when projectile is at highest point,

Δ = u(1-cosθ)= 2usin²θ/2

- If two projectile are projected with same speed at different angles then for same speed at different angles then for same range,

i. Sum of angles of projection must be 90°

θ + α = 90°,

ii. H_θ/H_α = tan²θ or cot²α

iii. (t_f)_θ/(t_f)_α = tanθ = cotα

iv. R = ½g(t_f)_θ(t_f)_α ⇒ R = ½gt₁t₂:t₁t₂ ∝ R

v. R = 4√( H_θH_α) ⇒ R²∝ H₁H₂

vi. H₁+ H₂= u²/2g.

Horizontal projection from Height :

For vertical motion

y = ½gt²…………………….(i)

for horizontal motion

t = x/u ……………….(ii)

from equation (i) and (ii),

y = gx²/2u²

y = (g/2u²).x² is the equation of parabola.

For time of flight :

H = ½gT²

T = √(2h/g)

For horizontal range :

Range (R) = u.T = u√(2h/g)

Some important tips :

When a ball rolled off from top of staircase with horizontal velocity ‘u’ having width ‘b’ and height ‘h’ the ball hits n^th step then, n =

2hu²

gb²

- If a man hits the target, he should point his gun in a direction higher than the target.

- If a man fires his gun directly aimed towards monkey at height, at same instant monkey at height, at same instant monkey starts falling then bullet hits the monkey.

- If two bodies are projected horizontally from certain height with different velocities u₁ and u₂ in opposite direction then

i. Their velocities are perpendicular after time,

ii. Velocities of 1^st body and 2^nd body when their velocity are perpendicular,v₁ =√(u₁²+ u₁u₂) and v₂ = √(u₂²+ u₁u₂)

iii. Their position vectors are perpendicular of after time,

t =

2√(u₁u₂)

g

- If a ball is droped from height ‘h’ from the top of an inclined plane of inclination ‘α’, ball elastically collides the it again, strike the inclined plane at a distance.

S = 8hSinα.