Question

A badminton shuttlecock is struck by a racquet at 45 degree angle, giving it an inital speed of 15m/s. How high will the shuttlecock go? How far will it travel horizontally before being contacted by the opponent's racquet at the same height from which it was projected?

Vertical component:

Answer 1

The maximum height the shuttlecock will reach can be calculated using the equation S = (usin(45))^2/9.8. The horizontal distance the shuttlecock will travel before being contacted by the opponent's racquet can be calculated using the equation S = ucos(45) * 2(usin(45))/9.8.

Step-by-step explanation:

The first step is to analyze the vertical component of the motion. We can use the equation v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken. The initial velocity in the vertical direction is given by u = usin(45), where usin(45) is the vertical component of the initial velocity. The final velocity in the vertical direction is zero at the maximum height. Plugging in the values, we get 0 = usin(45) - 9.8t.

Using the equation, we can solve for t = usin(45)/9.8.

The time taken to reach the maximum height is the same as the time taken to come back down to the same height. Therefore, the total time of flight is 2t. Plugging in the values, we get the total time of flight = 2(usin(45))/9.8.

The vertical displacement at the maximum height can be found using the equation S = ut + 0.5gt^2. Plugging in the values, we get S = (usin(45))^2/9.8.

The horizontal component of the motion is not affected by gravity. Therefore, the horizontal displacement can be calculated using the equation S = ut, where u is the initial velocity in the horizontal direction and t is the total time of flight. Plugging in the values, we get S = ucos(45) * 2(usin(45))/9.8.