To find the net vertical displacement of the baseball during its trajectory, we can use the kinematic equations of motion.
The initial velocity of the baseball can be resolved into horizontal and vertical components:
Viy = 29.4 sin(40) = 18.9 m/s (upward)
Vix = 29.4 cos(40) = 22.6 m/s (horizontal)
The total time the baseball is in the air is 3 seconds. We can use this to find the total vertical displacement using the following kinematic equation:
Δy = Viy*t + 0.5*a*t^2
where a is the acceleration due to gravity (-9.8 m/s^2).
Plugging in the values:
Δy = 18.9*3 + 0.5*(-9.8)*(3)^2
Δy = 56.7 - 44.1
Δy = 12.6 m
Therefore, the net vertical displacement of the baseball during its trajectory is 12.6 meters. Thus, the closest option is 11.0 m, which is the final answer.
A baseball thrown from the outfield is released from shoulder heigh at an initial velocity of 29.4 m/s at an initial angle of angle of 40.0° with respect to the horizontal. If it is in trajectory for a total of 3.00s before being caught by the third baseman at an equal shoulder-height level, what is the ball's net vertical displacement during its
3
S
trajectory?
9.80 m
•0 m
44.1 m
• 22.1 m
11.0 m
1 answer
1 answer