A baseball pitcher throws a baseball horizontally at a linear speed of 50.6 m/s. Before being caught, the baseball travels a horizontal distance of 24.9 m and rotates through an angle of 54.5 rad. The baseball has a radius of 3.43 cm and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball?

Question

I'm confused as to how to start this problem.

Answer 1

Quite a few assumptions have to be made as the question does not clarify them, and left them as "taken for granted".
1. The tangential speed is required at the moment the ball was caught.
2. the rotation of the ball was about an axis perpendicular to the direction of flight.
3. the maximum tangential speed is sought, namely at a point on the "equator" of the ball that is in the same direction as the translational motion of the ball.
4. All interaction with air (resistance) is neglected.

With the above assumptions in mind, we can proceed as follows:

1. using kinematics equations, find the duration of flight,t, using
Δx=vx&(t)
where
vx=50.6 m/s
Δx=24.9 m
both given in the question.
2. Find the vertical velocity (downwards, therefore negative) at time t. Since ball is subject to gravity,
vy=0+(1/2)gt²
vy=vertical velocity
t=time found in (1) above
g=acceleration due to gravity=9.8 m/s²
3. Find rotation velocity, ω
ω=54.5 radians /t
4. Find tangential speed due to rotation
vt=rω
where
r=radius of ball
= 3.43 cm
= 0.0343 m
5. Find total tangential speed at time t
At the moment the ball was caught, the ball was travelling at vx horizontally, vy vertically and vt tangentially.
vx and vy can be combined using Pythagoras theorem. The resulting maximum velocity V can then be combined with the tangential velocity by scalar addition.
V=√(vx²+vy²)+vt

Work through the steps, and post if you have questions or if clarifications are required.

Answer 2

*
Δx=(vx)(t)
g=acceleration due to gravity=-9.8 m/s²

Answer 3

j