A boy thrown a ball into air at 45^circ$45^{\circ}$ from the horizontal to land it on a roof of a building of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of H is ____ m. (g = 10 text m/s^2$g = 10 \text{ m/s}^{2}$)

### Related Formula u_x = u costheta = v cosphi$$u_x = u \cos\theta = v \cos\phi$$ ### Core Logic

Since horizontal component of velocity remains constant: u cos 60^circ = 20 cos 45^circ$$u \cos 60^{\circ} = 20 \cos 45^{\circ}$$ fracu2 = frac20sqrt2$$\frac{u}{2} = \frac{20}{\sqrt{2}}$$ u = frac40sqrt2 = 20sqrt2 text m/s$$u = \frac{40}{\sqrt{2}} = 20\sqrt{2} \text{ m/s}$$ ### Pattern Recognition Sees: Projectile motion speed at intermediate angle. Shortcut: Equate horizontal velocity components before and during flight. Check: Matches option (4). ✓ ### Chapter Mix Class 11 Physics: Motion in a Plane

### Related Formula K = frac12mu^2$$K = \frac{1}{2}mu^{2}$$ K_x = frac12m(u cos theta)^2 = K cos^2 theta$$K_{x} = \frac{1}{2}m(u \cos \theta)^{2} = K \cos^{2} \theta$$ ### Core Logic For a projectile with no air resistance, horizontal velocity remains constant. At the highest point (B), the vertical velocity is zero, so the kinetic energy is entirely due to the horizontal velocity. At point C, it hits the ground with the same speed as projected, restoring full kinetic energy. ### Step 1: Find KE at distinct points (textKE)_A = K = frac12mu^2$$(\text{KE})_{A} = K = \frac{1}{2}mu^{2}$$ At maximum height (B), velocity is v_B = u cos 60^circ = fracu2$v_{B} = u \cos 60^{\circ} = \frac{u}{2}$. (textKE)_B = frac12mleft(fracu2right)^2 = frac14left(frac12mu^2right) = fracK4$$(\text{KE})_{B} = \frac{1}{2}m\left(\frac{u}{2}\right)^{2} = \frac{1}{4}\left(\frac{1}{2}mu^{2}\right) = \frac{K}{4}$$ At landing point (C), speed is identical to initial speed u$u$. (textKE)_C = K$$(\text{KE})_{C} = K$$ ### Step 2: Calculate Ratio The required ratio is between the difference in kinetic energies at points C and B, to that at point A: textRatio = fracK_C - K_BK_A$$\text{Ratio} = \frac{K_{C} - K_{B}}{K_{A}}$$ textRatio = fracK - fracK4K = fracfrac3K4K = frac34$$\text{Ratio} = \frac{K - \frac{K}{4}}{K} = \frac{\frac{3K}{4}}{K} = \frac{3}{4}$$ ### Pattern Recognition Sees: "KE at highest point" → Substitute K cos^2 theta$K \cos^2 \theta$. For theta = 60^circ$\theta = 60^{\circ}$, cos 60^circ = 1/2$\cos 60^{\circ} = 1/2$, so K_B = K/4$K_B = K/4$. The difference with ground is 3K/4$3K/4$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane

### Related Formula For a projectile with launch speed u$u$ and angle theta$\theta$: - Horizontal Range: R = fracu^2 sin(2theta)g = frac2 u^2 sintheta costhetag$$R = \frac{u^2 \sin(2\theta)}{g} = \frac{2 u^2 \sin\theta \cos\theta}{g}$$ - Maximum Height: H = fracu^2 sin^2theta2g$$H = \frac{u^2 \sin^2\theta}{2g}$$ The general ratio linking range and maximum height is: tantheta = frac4HR$$\tan\theta = \frac{4H}{R}$$ ### Core Logic Given state: R = 3H Rightarrow fracHR = frac13$$R = 3H \Rightarrow \frac{H}{R} = \frac{1}{3}$$ ### Step 1: Determine the projection angle (theta$\theta$) Substitute the ratio into the relation: tantheta = 4 left(fracHRright) = 4 left(frac13right) = frac43$$\tan\theta = 4 \left(\frac{H}{R}\right) = 4 \left(\frac{1}{3}\right) = \frac{4}{3}$$ This is a standard Pythagorean triangle angle: sintheta = frac45, quad costheta = frac35$$\sin\theta = \frac{4}{5}, \quad \cos\theta = \frac{3}{5}$$ ### Step 2: Compute the horizontal range (R$R$) R = frac2 u^2 sintheta costhetag$$R = \frac{2 u^2 \sin\theta \cos\theta}{g}$$ R = frac2 u^2 left(frac45right) left(frac35right)g = frac24 u^225 g$$R = \frac{2 u^2 \left(\frac{4}{5}\right) \left(\frac{3}{5}\right)}{g} = \frac{24 u^2}{25 g}$$ Comparing this with the given format fracnu^225g$\frac{nu^2}{25g}$: n = 24$n = 24$ ### Pattern Recognition The relation tantheta = 4H/R$\tan\theta = 4H/R$ is an essential identity in projectile dynamics. Whenever R = k H$R = k H$, then tantheta = 4/k$\tan\theta = 4/k$. Recognizing standard angles like tantheta = 4/3$\tan\theta = 4/3$ or 3/4$3/4$ directly yields trigonometric values immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane

### Related Formula The time of flight T$T$ of a projectile launched with speed u$u$ at an angle theta$\theta$ with the horizontal is: T = frac2u sinthetag$$T = \frac{2u \sin\theta}{g}$$ ### Core Logic The launch angles of the two projectiles are: $ ### Core Logic The launch angles of the two projectiles are: $theta_1 = 45^circ + alpha$\theta_1 = 45^\circ + \alpha$ $ $theta_2 = 45^circ - alpha$\theta_2 = 45^\circ - \alpha$ Since they have the same speed $ Since they have the same speed $u: $: $fracT_1T_2 = fracsin(45^circ + alpha)sin(45^circ - alpha)$\frac{T_1}{T_2} = \frac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)}$ ### Step 1: Simplify Trigonometric Ratio Using the angle sum and difference formulas: $ ### Step 1: Simplify Trigonometric Ratio Using the angle sum and difference formulas: $fracT_1T_2 = fracsin 45^circ cos alpha + cos 45^circ sin alphasin 45^circ cos alpha - cos 45^circ sin alpha$\frac{T_1}{T_2} = \frac{\sin 45^\circ \cos \alpha + \cos 45^\circ \sin \alpha}{\sin 45^\circ \cos \alpha - \cos 45^\circ \sin \alpha}$ $ $fracT_1T_2 = fracfrac1sqrt2cosalpha + frac1sqrt2sinalphafrac1sqrt2cosalpha - frac1sqrt2sinalpha = fraccosalpha + sinalphacosalpha - sinalpha$\frac{T_1}{T_2} = \frac{\frac{1}{\sqrt{2}}\cos\alpha + \frac{1}{\sqrt{2}}\sin\alpha}{\frac{1}{\sqrt{2}}\cos\alpha - \frac{1}{\sqrt{2}}\sin\alpha} = \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha}$ Divide numerator and denominator by $ Divide numerator and denominator by $\cos\alpha: $: $fracT_1T_2 = frac1 + tanalpha1 - tanalpha$\frac{T_1}{T_2} = \frac{1 + \tan\alpha}{1 - \tan\alpha}$$ ### Pattern Recognition Sees: Projectile angles complementary to $ ### Pattern Recognition Sees: Projectile angles complementary to $45^\circ. Shortcut: Remember the identity $. Shortcut: Remember the identity $\tan(45^\circ + \alpha) = \frac{1+\tan\alpha}{1-\tan\alpha}. Since complementary angles have sine ratios proportional to $. Since complementary angles have sine ratios proportional to $\sin(45^\circ + \alpha)/\sin(45^\circ - \alpha) = \tan(45^\circ + \alpha), the answer is directly $, the answer is directly $\frac{1+\tan\alpha}{1-\tan\alpha}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane