A shuttle cork hitted upward from badminton racket with a velocity of ...
Problem:
A shuttle cork is hit upward from a badminton racket with a velocity of 50 m/s and it reaches 3 m from the hitting point in the last second of its upward journey. If the same shuttle cork is hit upward with a velocity of 200 m/s, then what will be the distance travelled in the last second of its upward journey?
Solution:
Step 1: Calculate the time taken by the shuttle cork to reach the maximum height.
When a shuttle cork is hit upward, it follows a projectile motion. The time taken by the shuttle cork to reach the maximum height can be calculated using the following formula:
Time taken to reach the maximum height, t = u/g where,
u = initial velocity = 50 m/s
g = acceleration due to gravity = 9.8 m/s²
Therefore, t = 50/9.8 = 5.1 seconds.
Step 2: Calculate the maximum height reached by the shuttle cork.
The maximum height reached by the shuttle cork can be calculated using the following formula:
Maximum height, h = u²/2g where,
u = initial velocity = 50 m/s
g = acceleration due to gravity = 9.8 m/s²
Therefore, h = 50²/2 x 9.8 = 127.55 meters.
Step 3: Calculate the velocity of the shuttle cork at the maximum height.
The velocity of the shuttle cork at the maximum height can be calculated using the following formula:
Velocity at maximum height, v = 0 m/s (as the shuttle cork stops at the maximum height)
Step 4: Calculate the time taken by the shuttle cork to reach 3 meters from the hitting point.
The time taken by the shuttle cork to reach 3 meters from the hitting point can be calculated using the following formula:
Distance travelled in the last second, s = 3 meters
Time taken to travel the last second, t1 = 1 second
Time taken to travel the remaining distance, t2 = t - t1 = 5.1 - 1 = 4.1 seconds
Therefore, the distance travelled by the shuttle cork in the last 4.1