zerragonoss said:
mysecondlife said:
A baseball is thrown at 40.0 m/s. Mike Trout hits it back in the opposite direction at a speed of 65.0 m/s. The ball is in contact with the bat for 1.20ms.
A)Find the acceleration of the ball assuming it is constant
B)the distance the ball travels while it is in contact with the bat.
K thanks. I thought you'd never ask!
A= (V1-V2)/T V1= 40 m/s V2= -65 m/s T=1.2x10^-3s
A= 8.75x10^4 m/s/s
D=1/2AT^2
D=2.52m x10^-2
Acceleration is correct with a=87500 m/s^2
Distance is not, as d=1/2*a*t^2 assumes starting from rest. Instead find deceleration and acceleration time separately:
t_1=v_1/a=0.457 ms (milliseconds)
t_2=v_2/a=0.743 ms
Then determine distance traveled in each direction:
d_1=0.5*a*t_1^2=9 mm (millimeters)
d_2=24 mm
Total distance moved is 33 mm. Ball stops touching bat 15 mm closer to the pitcher than when started touching bat.
For OP: How much do you care about physics? What do you think about Schrodinger's cat?