John throws a baseball from the outfield from shoulder height,at an initial velocity of 29.4 m/s at an initial angle of 30.0 $\degree$ with respect to the horizontal.The ball is in its trajectory for a total interval of 3.00 s before the third baseman catches it at an equal shoulder-height level.(Assume air resistance negligible. )What is the ball's horizontal displacement?
A)76.4 m
B)38.2 m
C)57.3 m
D)zero

Question

Quizplus · Accepted Answer

The horizontal displacement can be calculated using the formula: horizontal velocity = initial velocity * cos(angle). The horizontal velocity is 29.4 m/s * cos(30°) = 25.5 m/s. The horizontal displacement is then horizontal velocity * time = 25.5 m/s * 3.00 s = 76.5 m, which is approximately 76.4 m.

John Throws a Baseball from the Outfield from Shoulder Height,at °\degree°

John Throws a Baseball from the Outfield from Shoulder Height,at $\degree$