Question

A body is projected up with a velocity of $30 \text{ ms}^{-1}$ at an angle of $30^\circ$. The ratio of maximum height reached to the height reached in the first second is ($g = 10 \text{ms}^{-2}$)}

• A. $10 : 9$
• B. $10 : 8$
• C. $9 : 8$
• D. $9 : 5$
• E. $5 : 4$

Hint:

To simplify ratios like 11.25 : 10, write it as a fraction: $\frac{1125}{1000}$. Successive division by 5 or identifying 125 as a common factor makes it very quick.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a 2D projectile motion problem. The height refers to the vertical displacement. We need to calculate the maximum vertical height achieved and the vertical displacement at exactly \(t = 1\) second, then find their ratio.
Step 2: Key Formula or Approach:
Initial vertical velocity: \(u_y = u \sin \theta\).
Maximum height: \(H = \frac{u_y^2}{2g} = \frac{u^2 \sin^2 \theta}{2g}\).
Height at time \(t\): \(y(t) = u_y t - \frac{1}{2} gt^2\).
Step 3: Detailed Explanation:
Given \(u = 30\text{ m/s}\), \(\theta = 30^\circ\), \(g = 10\text{ m/s}^2\).
Calculate the initial vertical velocity:
\[ u_y = 30 \sin 30^\circ = 30 \left(\frac{1}{2}\right) = 15\text{ m/s} \] Calculate the maximum height \(H\):
\[ H = \frac{u_y^2}{2g} = \frac{15^2}{2(10)} = \frac{225}{20} = 11.25\text{ m} \] Calculate the height \(h\) reached in the first second (\(t = 1\)):
\[ h = y(1) = 15(1) - \frac{1}{2}(10)(1)^2 \] \[ h = 15 - 5 = 10\text{ m} \] Find the ratio \(H : h\):
\[ \text{Ratio} = \frac{11.25}{10} = 1.125 \] Convert to a fraction by multiplying numerator and denominator by 1000:
\[ \frac{1125}{1000} \] Divide both by 125:
\[ 1125 \div 125 = 9 \] \[ 1000 \div 125 = 8 \] The ratio is \(9/8\) or \(9 : 8\).
Step 4: Final Answer:
The ratio is 9 : 8.