Question
An outfielder throws a .150 kg baseball at a speed of 40.0 m/s and an initial angle of 30 degrees. What is the kinetic energy of the ball at the highest point of its motion.
Work:
0^2=40^2+(.002)(-9.8m/s^2)(d)
d=81.6m
W=(.150kg)(9.8m/s^2)(cos30)(81.6m)
=104J
I got the wrong answer though. The answer is suppose to be 90J.
Work:
0^2=40^2+(.002)(-9.8m/s^2)(d)
d=81.6m
W=(.150kg)(9.8m/s^2)(cos30)(81.6m)
=104J
I got the wrong answer though. The answer is suppose to be 90J.
Answers
Raj
to find the vertical velocity we use
40*cos(30)=34.6 m/s
W= KE = 1/2 mv^2
=1/2*0.15*34.6^2= 89.797 J or 90 J
40*cos(30)=34.6 m/s
W= KE = 1/2 mv^2
=1/2*0.15*34.6^2= 89.797 J or 90 J