TG EAMCET 3 May 2025 Shift 1 Paper

Step 1: Write the formula for maximum heightThe formula for the highest point a projectile can reach is:$H = \frac{u^2 \sin^2 \theta}{2g}$where $H$ is maximum height, $u$ is the starting speed, $\theta$ is the angle of projection, and $g$ is gravity.Step 2: Put in the given valuesWe know $H = 9.8\ \mathrm{m},\ u = 19.6\ \mathrm{m\,s^{-1}}$, and $g = 9.8\ \mathrm{m\,s^{-2}}$. We put these in the formula:$9.8 = \frac{(19.6)^2 \sin^2 \theta}{2 \times 9.8}$Step 3: Simplify to find $\sin^2 \theta$Multiply both sides by $2 \times 9.8$ :$9.8 = 19.6 \times 19.6 \times \frac{\sin^2 \theta}{2 \times 9.8}$$\Rightarrow 9.8 = 39.2 \sin^2 \theta$Divide both sides by 19.6,$\sin^2 \theta = \frac{1}{2}$Step 4: Find the angle $\theta$$\sin \theta = \frac{1}{\sqrt{2}} = \sin 45^{\circ},\ \text{so}$$\theta = 45^{\circ}.$Step 5: Write the formula for range $R$The formula for how far the projectile lands (range) is: $R = \frac{u^2 \sin 2\theta}{g}$Step 6: Put in the numbersSince $u = 19.6\ \mathrm{m\,s^{-1}},\ \theta = 45^{\circ}$, and $g = 9.8\ \mathrm{m\,s^{-2}}$,$R = \frac{(19.6)^2 \times \sin (2 \times 45^{\circ})}{9.8}$$\sin (90^{\circ}) = 1,\ \text{so}$$R = \frac{(19.6)^2 \times 1}{9.8}$$R = \frac{384.16}{9.8} = 39.2\ \mathrm{m}$