Answer :
To find the horizontal component of the velocity of the arrow, we need to understand the concepts of projectile motion. When an object is projected into the air and moves under the influence of gravity, it has both a horizontal and vertical component of motion.
In this case, we are given:
- The range (horizontal distance) [tex]\( R \)[/tex] is 200 meters.
- The time of flight [tex]\( t \)[/tex] is 5 seconds.
The horizontal component of the velocity [tex]\( V_x \)[/tex] is constant throughout the flight of the projectile because there are no horizontal forces acting on it (assuming air resistance is negligible). We can find the horizontal velocity by dividing the range by the time of flight.
Step-by-step solution:
1. Identify the given values:
- Range (R): 200 meters
- Time of flight (t): 5 seconds
2. Recall the formula for the horizontal component of the velocity for a projectile:
[tex]\[ V_x = \frac{R}{t} \][/tex]
where [tex]\( V_x \)[/tex] is the horizontal component of the velocity, [tex]\( R \)[/tex] is the range, and [tex]\( t \)[/tex] is the time of flight.
3. Substitute the given values into the formula:
[tex]\[ V_x = \frac{200 \text{ meters}}{5 \text{ seconds}} \][/tex]
4. Perform the division:
[tex]\[ V_x = 40 \text{ meters/second} \][/tex]
Therefore, the horizontal component of the velocity of the arrow is [tex]\( 40 \text{ meters/second} \)[/tex].
In this case, we are given:
- The range (horizontal distance) [tex]\( R \)[/tex] is 200 meters.
- The time of flight [tex]\( t \)[/tex] is 5 seconds.
The horizontal component of the velocity [tex]\( V_x \)[/tex] is constant throughout the flight of the projectile because there are no horizontal forces acting on it (assuming air resistance is negligible). We can find the horizontal velocity by dividing the range by the time of flight.
Step-by-step solution:
1. Identify the given values:
- Range (R): 200 meters
- Time of flight (t): 5 seconds
2. Recall the formula for the horizontal component of the velocity for a projectile:
[tex]\[ V_x = \frac{R}{t} \][/tex]
where [tex]\( V_x \)[/tex] is the horizontal component of the velocity, [tex]\( R \)[/tex] is the range, and [tex]\( t \)[/tex] is the time of flight.
3. Substitute the given values into the formula:
[tex]\[ V_x = \frac{200 \text{ meters}}{5 \text{ seconds}} \][/tex]
4. Perform the division:
[tex]\[ V_x = 40 \text{ meters/second} \][/tex]
Therefore, the horizontal component of the velocity of the arrow is [tex]\( 40 \text{ meters/second} \)[/tex].
