A baseball is struck by a batter at an angle such that it reaches its maximum height at the location of the pitcher-and just grazes the top of the pitcher's head (at an elevation of 1.50 m above the launch point of the baseball). The pitcher is a horizontal distance of 20.0 m from the point at which the ball was struck. What was the launch speed for the baseball?
A) 36.6 m/s
B) 36.1 m/s
C) 41.6 m/s
D) 30.7 m/s

Question

Quizplus · Accepted Answer

The launch speed can be found using the equations of motion under gravity. Given that the maximum height is 1.50 m and the horizontal distance is 20.0 m, we can use these values to calculate the initial velocity. The time to reach the maximum height is found using $t = \sqrt{\frac{2h}{g}}$, where $h$ is the height and $g$ is the acceleration due to gravity (9.8 m/s$^2$). The total time of flight is double this time since the time to go up is the same as the time to come down. The horizontal velocity ($v_x$) is constant, and using the total time of flight, we can find $v_x = \frac{d}{t}$, where $d$ is the horizontal distance. The initial velocity ($v_0$) can then be calculated knowing the vertical component of the velocity ($v_y$) at launch, which is related to the maximum height. The correct calculation leads to an initial velocity of approximately 36.6 m/s.

A Baseball Is Struck by a Batter at an Angle