Question

A body falling freely under gravity from rest from a certain height reaches the ground in \(5\) seconds. The distance travelled by the body in the last two seconds of its motion is:

• A. \(98 \, {m}\)
• B. \(44.1 \, {m}\)
• C. \(58.8 \, {m}\)
• D. \(78.4 \, {m}\)

Hint:

For objects in free fall, remember the distance increases quadratically with time due to the constant acceleration of gravity.

Answer 1

The Correct Option is D

Step 1: Determine the total distance fallen in 5 seconds. 
Using the formula for distance travelled under gravity: \[ s = \frac{1}{2}gt^2 = \frac{1}{2}(9.8 \, {m/s}^2)(5^2) = 122.5 \, {m} \] Step 2: Determine the distance fallen in the first 3 seconds. \[ s_1 = \frac{1}{2}(9.8 \, {m/s}^2)(3^2) = 44.1 \, {m} \] Step 3: Calculate the distance in the last two seconds. \[ s_2 = s - s_1 = 122.5 \, {m} - 44.1 \, {m} = 78.4 \, {m} \]

Answer 2

To solve the problem, we need to calculate the distance traveled by a body falling freely under gravity in the last two seconds of its motion.

1. Understanding the Problem:
The body falls freely under gravity, meaning its motion follows the equations of uniformly accelerated motion. The total time given is 5 seconds, and we are interested in the distance traveled during the last 2 seconds of this motion.

2. Using the Kinematic Equation:
The kinematic equation for distance traveled under uniform acceleration (gravity) is:

\[ s = ut + \frac{1}{2} a t^2 \] where: - \( s \) is the distance traveled, - \( u \) is the initial velocity (which is 0 for an object falling from rest), - \( a \) is the acceleration due to gravity (\( g = 9.8 \, \text{m/s}^2 \)), - \( t \) is the time. For the total time of 5 seconds, the distance traveled is: \[ s_5 = 0 \cdot 5 + \frac{1}{2} \times 9.8 \times 5^2 = \frac{1}{2} \times 9.8 \times 25 = 122.5 \, \text{m} \] For the first 3 seconds, the distance traveled is: \[ s_3 = 0 \cdot 3 + \frac{1}{2} \times 9.8 \times 3^2 = \frac{1}{2} \times 9.8 \times 9 = 44.1 \, \text{m} \] Thus, the distance traveled in the last 2 seconds is: \[ s_{\text{last 2}} = s_5 - s_3 = 122.5 - 44.1 = 78.4 \, \text{m} \]

3. Conclusion:
The distance traveled by the body in the last 2 seconds of its motion is 78.4 meters.

Final Answer:
The correct answer is (D) 78.4 m.