Sure, let's solve this step-by-step.
When a body is thrown vertically upwards, it will decelerate under the influence of gravity until it comes to a stop at the peak of its trajectory. After that, it will accelerate back downwards, again under the influence of gravity.
Given:
- Initial velocity, [tex]\( u = 50 \)[/tex] meters per second
- Acceleration due to gravity, [tex]\( g = 10 \)[/tex] meters per second squared (downwards)
### Step 1: Time to Reach the Highest Point
First, we need to calculate the time it takes for the body to reach the highest point of its trajectory. At the highest point, the final velocity ([tex]\( v \)[/tex]) will be [tex]\( 0 \)[/tex] meters per second.
We can use the following kinematic equation:
[tex]\[ v = u - gt \][/tex]
At the highest point, [tex]\( v = 0 \)[/tex]:
[tex]\[ 0 = 50 - 10t \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ 10t = 50 \][/tex]
[tex]\[ t = 5 \][/tex] seconds
Thus, the time to reach the highest point is [tex]\( 5 \)[/tex] seconds.
### Step 2: Total Time to Return to the Initial Position
The time to return to the initial position from the highest point will be the same as the time taken to reach the highest point (due to symmetry in projectile motion). Hence:
[tex]\[ \text{Total time} = 2 \times \text{Time to peak} \][/tex]
[tex]\[ \text{Total time} = 2 \times 5 \][/tex]
[tex]\[ \text{Total time} = 10 \][/tex] seconds
Therefore, the body will take [tex]\( 10 \)[/tex] seconds to return to the initial position.
Among the given options:
- a) 15 seconds
- b) 4 seconds
- c) 10 seconds
- d) 54 seconds
The correct answer is:
c) 10 seconds
