Answer:
Yes
Explanation:
The given parameters are;
The speed with which the fastball is hit, u = 49.1 m/s (109.9 mph)
The angle in which the fastball is hit, θ = 22°
The distance of the field = 96 m (315 ft)
The range of the projectile motion of the fastball is given by the following formula

Where;
g = The acceleration due to gravity = 9.81 m/s², we have;

Yes, given that the ball's range is larger than the extent of the field, the batter is able to safely reach home.
The impulse given to the ball is equal to the change in its momentum:
J = ∆p = (0.50 kg) (5.6 m/s - 0) = 2.8 kg•m/s
This is also equal to the product of the average force and the time interval ∆t :
J = F(ave) ∆t
so that if F(ave) = 200 N, then
∆t = J / F(ave) = (2.8 kg•m/s) / (200 N) = 0.014 s
Prototype
Chemistry
Technology
Troubleshooting
Hypothesis?
Observing
Variable
Hypothesis?
Engineer
System
High density
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