Question

If a body projected with a velocity of 19.6 ms⁻¹ reaches a maximum height of 9.8 m, then the range of the projectile is (Neglect air resistance)

• A. 19.6 m
• B. 39.2 m
• C. 78.4 m
• D. 9.8 m

Hint:

In projectile motion problems, recognize that values like 19.6, 9.8, and 4.9 are often multiples or factors of the acceleration due to gravity, g. Here, \(19.6 = 2g\) and \(9.8 = g\), which simplifies calculations significantly. Also, an angle of \(45^\circ\) gives the maximum possible range for a given initial speed.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a projectile motion problem. We are given the initial velocity and maximum height and need to find the horizontal range. We can use the formula for maximum height to find the angle of projection first, and then use that angle to find the range. We'll use \(g = 9.8 \text{ m/s}^2\).
Step 2: Key Formulae:
Maximum Height: \(H = \frac{u^2 \sin^2 \theta}{2g}\)
Horizontal Range: \(R = \frac{u^2 \sin(2\theta)}{g}\)
Given: \(u = 19.6\) m/s, \(H = 9.8\) m.
Step 3: Detailed Explanation:
First, we find the angle of projection, \(\theta\), using the maximum height formula.
\[ 9.8 = \frac{(19.6)^2 \sin^2 \theta}{2 \times 9.8} \] \[ (9.8) \times (2 \times 9.8) = (19.6)^2 \sin^2 \theta \] Since \(19.6 = 2 \times 9.8\), we can write:
\[ 2 \times (9.8)^2 = (2 \times 9.8)^2 \sin^2 \theta \] \[ 2 \times (9.8)^2 = 4 \times (9.8)^2 \sin^2 \theta \] \[ \sin^2 \theta = \frac{2 \times (9.8)^2}{4 \times (9.8)^2} = \frac{1}{2} \] \[ \sin \theta = \frac{1}{\sqrt{2}} \] This implies that the angle of projection is \(\theta = 45^\circ\).
Now, we can calculate the range \(R\) using the range formula and \(\theta = 45^\circ\).
\[ R = \frac{u^2 \sin(2\theta)}{g} = \frac{(19.6)^2 \sin(2 \times 45^\circ)}{9.8} \] \[ R = \frac{(19.6)^2 \sin(90^\circ)}{9.8} \] Since \(\sin(90^\circ) = 1\):
\[ R = \frac{(19.6)^2}{9.8} = \frac{19.6 \times 19.6}{9.8} = 2 \times 19.6 = 39.2 \text{ m} \] Step 4: Final Answer:
The range of the projectile is 39.2 m. Therefore, option (B) is correct.