top of page

Roger Federer

Acceleration of Racquet Take Back

 

Given:

vf = 4.36 m/s

vi = 2.31 m/s

t = 0.50 s

 

Required: 

a (acceleration)

Analyze: 

 

 

 

 

Solve: 

a = (4.36 m/s - 2.31 m/s)/0.50 s

    = 4.10 m/s^2

 

Present: 

Therefore, Roger Federer's takeback speed is 4.10 m/s^2

Acceleration of Racquet

Given:

vf = 45.62 m/s

vi = 38.52 m/s

t = 0.10 s 

 

Required: 

a (acceleration)

Analyze: 

 

 

 

 

Solve: 

a = (45.62 m/s - 38.52 m/s)/0.10 s

    = 71.00 m/s^2

 

Present: 

Therefore, the acceleration of the ball that Roger Federer hits on his forehand is 71.00 m/s^2

Given:

vf = 18.83 m/s

vi = 9.05 m/s

t = 0.17 m/s

 

Required: 

a (acceleration)

Analyze: 

 

 

 

 

Solve: 

a = (18.83 m/s - 9.05 m/s)/0.17 s

    = 57.53 m/s^2

 

Present: 

Therefore, Roger Federer's racquet speed after his take back is 57.53 m/s^2

 

Acceleration of Ball

bottom of page