Projectile Motion Formulas

1. Projectile Motion:

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

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

(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 = \(\frac{2 u \sin \theta}{g}=\frac{2 u_{y}}{a_{y}}\)

(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 ╬╕ (remains same)

(g) Vertical component of velocity at any time v_y = u sin ╬╕ – 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 ╬▒ = \(\frac{\mathbf{v}_{\mathbf{y}}}{\mathbf{v}_{\mathbf{x}}}\)

2. General Result:

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

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 ╬▒ = \(\frac{v_{y}}{v_{x}}\)

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

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

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

Projectile thrown from an inclined plane

• \(\overrightarrow{\mathrm{a}}_{\mathrm{x}}\) = -g sin ╬╕₀\(\hat{\mathrm{i}}\)
• \(\overrightarrow{\mathrm{a}}_{\mathrm{y}}\) = -g cos ╬╕₀\(\hat{\mathrm{j}}\)
  
• \(\overrightarrow{\mathrm{u}}_{\mathrm{x}}\) = u cos (╬╕ – ╬╕₀)\(\hat{\mathrm{i}}\)
• \(\overrightarrow{\mathrm{u}}_{\mathrm{y}}\) = u sin (╬╕ – ╬╕₀)\(\hat{\mathrm{j}}\)

(a) Time of flight T = \(\frac{2 \mathrm{u}_{\mathrm{y}}}{\mathrm{a}_{\mathrm{y}}}=\frac{2 \mathrm{u} \sin \left(\theta-\theta_{0}\right)}{\mathrm{g} \cos \theta_{0}}\)

R = u cos (╬╕ – ╬╕₀) T – \(\frac{1}{2}\) g sin ╬╕₀.T²
R = \(\frac{2 u^{2} \sin \left(\theta-\theta_{0}\right) \cos \theta}{g \cos ^{2} \theta_{0}}\)

Important for R_max ╬╕ = \(\frac{\pi}{4}+\frac{\theta_{0}}{2}\) and R_max = \(\frac{u^{2}}{g\left(1+\sin \theta_{0}\right)}\).

Maximum height H_max = \(\frac{\mathrm{u}_{\mathrm{y}}^{2}}{2 \mathrm{a}_{\mathrm{y}}}=\frac{\mathrm{u}^{2} \sin ^{2}\left(\theta-\theta_{0}\right)}{2 \mathrm{g} \cos \theta_{0}}\)

• R = nH, ╬╕ = tan^-1\(\left(\frac{\mathrm{A}}{\mathrm{n}}\right)\)
• At t = \(\frac{u}{g \sin \theta}\), velocity are ΓèÑ and v = u cot ╬╕
• v_avg = \(\frac{u}{2} \sqrt{1+3 \cos ^{2} \theta}\) for half motion
• KE_hight = \(\frac{1}{2}\) m (u cos ╬╕)² = \(\frac{1}{2}\) mu²cos²╬╕ = K₀cos²╬╕.

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