Projectile Motion Formulas

1. Projectile Motion:

Thrown at an angle θ with horizontal
${\overrightarrow{\mathrm{u}}}_{\mathrm{x}}$ = u cos θ $\hat{\mathrm{i}}$; a_x = 0
${\overrightarrow{\mathrm{u}}}_{\mathrm{y}}$ = u sin θ $\hat{\mathrm{j}}$; ${\overrightarrow{\mathrm{a}}}_{\mathrm{y}}=-\mathrm{g}\hat{\mathrm{j}}$

(a) y = x tan θ – $\frac{1}{2}\cdot g\cdot {\left[\frac{x}{u\cos \theta }\right]}^2$
or y = x tanθ $[1-\frac{x}{R}]$

(b) Time to reach maximum height (time of ascent/time of descent)
t = $\frac{u\sin \theta }{g}=\frac{u_y}{a_y}$

(c) Time of flight
T = $\frac{2u\sin \theta }{g}=\frac{2u_y}{a_y}$

(d) Horizontal range
R = $\frac{u^2\sin 2\theta }{g}$ = u_x × T

(e) Maximum height
H_max = $\frac{{\mathrm{u}}^2{\sin }^2\theta }{2\mathrm{g}}=\frac{{\mathrm{u}}_{\mathrm{y}}^2}{2{\mathrm{a}}_{\mathrm{y}}}$

(f) Horizontal velocity at any time v_x = u cos θ (remains same)

(g) Vertical component of velocity at any time v_y = u sin θ – gt

(h) Resultant velocity $\overrightarrow{\mathrm{v}}={\mathrm{v}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{v}}_{\mathrm{y}}\hat{\mathrm{j}}$
$\overrightarrow{\mathrm{v}}=u\cos \theta \hat{\mathrm{i}}+(u\sin \theta -\mathrm{gt})\hat{\mathrm{j}}$
$\mathrm{v}=|\overrightarrow{\mathrm{v}}|=\sqrt{{\mathrm{u}}^2+{\mathrm{g}}^2{\mathrm{t}}^2-2\mathrm{ugt}\sin \theta }$
and tan α = $\frac{{\mathbf{v}}_{\mathbf{y}}}{{\mathbf{v}}_{\mathbf{x}}}$

2. General Result:

• For maximum range θ = 45°, R_max = $\frac{u^2}{g}$
  and H_max = $\frac{R_{max}}{4}$, at θ = 45° and initial velocity u
  H_max = $\frac{R_{max}}{2}$, at θ = 90° and initial velocity u
• Change in momentum Δ$\overrightarrow{\mathrm{P}}$ = – mgt $\hat{j}$
• For complete motion = -2 m u sin θ $\hat{j}$
• At highest point = – m u sin θ $\hat{i}$

3. Projectile thrown parallel to the horizontal from height ‘h’

u_x = u v_x = u
u_y = 0 v_y = -gt (upward)
(a) Equation y = –$\frac{1}{2}g\frac{x^2}{u^2}$
(b) Velocity at any time
v = $\sqrt{u^2+g^2{t}^2}$
tan α = $\frac{v_y}{v_x}$

(c) Displacement S = $x\hat{i}+y\hat{j}=ut\hat{i}-\frac{1}{2}g{t}^2\hat{j}$

(d) Time of flight T = $\sqrt{\frac{2h}{g}}$

(e) Horizontal Range = u$\sqrt{\frac{2h}{g}}$ = u_x T

Projectile thrown from an inclined plane

• ${\overrightarrow{\mathrm{a}}}_{\mathrm{x}}$ = -g sin θ₀$\hat{\mathrm{i}}$
• ${\overrightarrow{\mathrm{a}}}_{\mathrm{y}}$ = -g cos θ₀$\hat{\mathrm{j}}$
  
• ${\overrightarrow{\mathrm{u}}}_{\mathrm{x}}$ = u cos (θ – θ₀)$\hat{\mathrm{i}}$
• ${\overrightarrow{\mathrm{u}}}_{\mathrm{y}}$ = u sin (θ – θ₀)$\hat{\mathrm{j}}$

(a) Time of flight T = $\frac{2{\mathrm{u}}_{\mathrm{y}}}{{\mathrm{a}}_{\mathrm{y}}}=\frac{2\mathrm{u}\sin (\theta -{\theta }_0)}{\mathrm{g}\cos {\theta }_0}$

R = u cos (θ – θ₀) T – $\frac{1}{2}$ g sin θ₀.T²
R = $\frac{2u^2\sin (\theta -{\theta }_0)\cos \theta }{g{\cos }^2{\theta }_0}$

Important for R_max θ = $\frac{\pi }{4}+\frac{{\theta }_0}{2}$ and R_max = $\frac{u^2}{g(1+\sin {\theta }_0)}$.

Maximum height H_max = $\frac{{\mathrm{u}}_{\mathrm{y}}^2}{2{\mathrm{a}}_{\mathrm{y}}}=\frac{{\mathrm{u}}^2{\sin }^2(\theta -{\theta }_0)}{2\mathrm{g}\cos {\theta }_0}$

• R = nH, θ = tan^-1$\left(\frac{\mathrm{A}}{\mathrm{n}}\right)$
• At t = $\frac{u}{g\sin \theta }$, velocity are ⊥ and v = u cot θ
• v_avg = $\frac{u}{2}\sqrt{1+3{\cos }^2\theta }$ for half motion
• KE_hight = $\frac{1}{2}$ m (u cos θ)² = $\frac{1}{2}$ mu²cos²θ = K₀cos²θ.

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