A tennis ball is dropped from a height of 2.0 m and it rebounds to a height of 1.5 m . If the ball is in contact with the floor for 0.110 s , what is the acceleration of the ball when it is in contact with the floor ?
I am a bit confused with the directions .
For the downward motion , g=-9.81m/s^2 , s=-2.0m and u =0
And when it rebounds which will be the upward motion , g=-9.81m/s^2 . s=1.5
That is what the book says .
For the downward motion , isnt the acceleration due to gravity supposed to be negative since it is in the direction of gravity and the displacement , s , why is it negative ??
And for the upward motion , why is the displacement positive ??
Ok .. so lets say i take downwards as the negative direction , do i say that the displacement , velocity and acceleration are all negative ?
But i thought the acceleration should be positive as it is in the direction of the gravity ?
But consider projectile motion , when an object is being projected at an angle from a building , lets say i take the upwards direction as positive , do i also say that the acceleration is -9.81m/s^2 throughout its motion ?
With this in mind and 350 years experience, take the point that the ball bounces as 0, and the upward direction as positive.
Then the ball starts from s=2, with acceleration -g (that is downwards, as we have defined + as upwards)
Now calculate the velocity when the ball hits the ground, and its velocity on rebound. The (avaerage) accelleration is the change in velocity divided by the time it takes.