Question

A projectile is projected with the velocity of \( 3\hat{i} + 4\hat{j} \, {m/s} \). The horizontal range of the projectile will be:

• A. 1.2 m
• B. 2.4 m
• C. 3.6 m
• D. 4.5 m

Hint:

To find the range of a projectile, use the horizontal component of velocity and the time of flight. The time of flight is determined by the vertical motion, and the horizontal range is the product of horizontal velocity and time of flight.

Answer 1

The Correct Option is B

Step 1: The given velocity components are: - Horizontal component of velocity: \( u_x = 3 \, {m/s} \), - Vertical component of velocity: \( u_y = 4 \, {m/s} \). Step 2: The time of flight for a projectile is given by the formula: \[ t = \frac{2 u_y}{g} \] where \( g = 9.8 \, {m/s}^2 \) is the acceleration due to gravity. Substituting the value of \( u_y \): \[ t = \frac{2 \times 4}{9.8} = \frac{8}{9.8} \approx 0.816 \, {seconds} \] Step 3: The horizontal range is given by: \[ R = u_x \times t \] Substituting the values of \( u_x \) and \( t \): \[ R = 3 \times 0.816 \approx 2.448 \, {m} \] Thus, the horizontal range is approximately \( 2.4 \, {m} \).