Projectile Motion NEET 2025 PDF

# ARJUNA NEET 2025 ## Lecture 03: समतल में गति **By - Vishnu Nagar (VN Sir)** **Physics** **Topics to be covered** - क्षेतिज प्रक्षेप्य गति - वृत्तीय गती ## समतल में गति **प्रक्षेप्य गति की समीकरण** - $y = xtan\theta - \frac{1}{2}gt^2/cos^2\theta$ - $y = xtan\theta - \frac{1}{2}gt^2(1+tan^2...

# ARJUNA NEET 2025 ## Lecture 03: समतल में गति **By - Vishnu Nagar (VN Sir)** **Physics** **Topics to be covered** - क्षेतिज प्रक्षेप्य गति - वृत्तीय गती ## समतल में गति **प्रक्षेप्य गति की समीकरण** - $y = xtan\theta - \frac{1}{2}gt^2/cos^2\theta$ - $y = xtan\theta - \frac{1}{2}gt^2(1+tan^2\theta)/tan^2\theta$ - $y = xtan\theta - tan^2\theta * x^2/R$ **Important Equations for Projectile motion** $H = U^2sin^2\theta/2g$ $H = U_y^2/2g$ $T = 2U_y/g = 2Usin\theta/g$ $R = U^2sin2\theta/g = 2UxUy/g$ $R_{max} = U^2/g$ **If $\theta = 45^{\circ}$** - $H = Rtan\theta/4$ - $T = \sqrt{8H/g}$ **पुरक कोणी पर प्रधगति** $R_1 = R_2 = R = U^2sin2\theta/g$ $R = 4\sqrt{H_1H_2}$ **पुरक कोर्णा में अन्य** $R_1 = R_2 = R = U^2sin2\theta/g$ $H_1 = U^2sin^2\theta/2g$ $H_2 = U^2cos^2\theta/2g$ $H_1 * H_2 = \frac{1}{4} * (\frac{U^2sin2\theta }{g})^2$ $H_1*H_2 = \frac{1}{4*4} * (\frac{U^2sin2\theta }{g})^2$ $H_1*H_2 = \frac{1}{16} (R^2)$ $16H_1H_2 = R^2$ $4\sqrt{H_1H_2} = R$ **Some Important Points** - क्षेतिज दिशा का वेग नियत रहता है - प्रक्षेप्य गति के दौरान Hmax पर प्रडिशा का रंग शुन्य होता है **Illustration** Consider a Body moving from point A to B in Projectile Path - $V_f = Ucos\theta \hat{i} + Usin\theta \hat{j}$ - $V_f = Ucos\theta \hat{i}$ - $\Delta V = V_f - V_i$ = -$Usin\theta \hat{j}$ - $|\Delta V| = Usin\theta$ **Numerical** A body is projected with an initial velocity of 50 m/s at an angle of 37 degrees to the horizontal. Find 1. **The magnitude of the velocity of the body at point P.** 2. **The time taken to reach point P.** **Solution** - Resolve the initial velocity into horizontal and vertical components. - Apply the equations of motion to find the vertical component of the velocity at point P. - Use the Pythagorean theorem to find the magnitude of the velocity at point P. **Shortcut** If a body is projected with an initial velocity of \[U] at an angle of \[\theta] to the horizontal, then the time taken to reach the maximum height is given by $t = \frac{U sin\theta }{g}$ **So the answer is:** $t = \frac{50}{10} = \frac{25}{3}$ ## क्षैतिज प्रक्षेप्य गति **x दिशा** - $U_x = U$ - $a_x = 0$ **y दिशा** - $U_y = 0$ - $a_y = -g$ **Important Equations for Horizontal Projectile Motion** - $S = Ut + 1/2at^2$ - $X = U_xt + 0$ - $X = U_xt$ **Equations for horizontal and vertical direction** 1.) $x = Ut = \frac{U*2h}{\sqrt{g}}$ 2.) $h = \frac{1}{2}gt^2$ ## **Numerical** A projectile is launched horizontally from the top of a cliff with an initial velocity of \[U] . The projectile lands at a distance \[x] from the base of the cliff. What is the height of the cliff? **Solution** Use the equation for the horizontal distance traveled x = Ut t = x --- U Then use the equation for the vertical distance traveled $h = \frac{1}{2}gt^2$ **Answer:** $h = \frac{1}{2}g\frac{x^2}{U^2}$ ## Vertical Projectile Motion **x दिशा** - $U_x = U$ - $a_x = 0$ **y दिशा** - $U_y = 0$ - $a_y = -g$ - $V = V_x\hat{i} + V_y\hat{j}$ - $V = U_i-gt\hat{j}$ - $V_y = -gt$ ## Numerical A body is thrown vertically upwards with an initial velocity of \[u]. What is the time taken by the body to reach the maximum height? **Solution** Use the equation for the final velocity of the body $V = U + at$ At maximum height $V = 0$ $0 = U -(g)t$ $t = U/g$ **Answer:** $t = U/g$ ## The final answer The time taken by the body to reach the maximum height is $t = U/g$.