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:
We use the standard kinematic equations for projectile motion to find the maximum height ($H$) and the vertical displacement ($h$) at a specific time $t$.

Step 2: Detailed Explanation:
Given $u = 30 \text{ m/s}$, $\theta = 30^\circ$, $g = 10 \text{ m/s}^2$.
1. Maximum Height reached ($H$):
\[ H = \frac{u^2 \sin^2 \theta}{2g} = \frac{30^2 \sin^2 30^\circ}{2(10)} \]
\[ H = \frac{900 \times (1/2)^2}{20} = \frac{45 \times (1/4)}{1} = 11.25 \text{ m} \]
2. Height reached in first second ($h$ at $t=1$):
The initial vertical velocity is $u_y = u \sin \theta = 30 \sin 30^\circ = 15 \text{ m/s}$.
Using $h = u_y t - \frac{1}{2}gt^2$:
\[ h = (15)(1) - \frac{1}{2}(10)(1)^2 = 15 - 5 = 10 \text{ m} \]
3. Required Ratio $H : h$:
\[ H : h = 11.25 : 10 \]
Multiply by 100 to remove decimals:
\[ 1125 : 1000 \]
Divide by 125:
\[ 9 : 8 \]

Step 3: Final Answer:
The ratio is $9 : 8$.