Final answer:
The speed of the baseball when it reaches the ground is approximately 20.9 m/s. The time of flight is approximately 1.9 seconds, and the vertical velocity when it reaches the ground is approximately -13.8 m/s.
Step-by-step explanation:
To find the speed of the baseball when it reaches the ground, we need to analyze its motion. Since the baseball is thrown horizontally, we can treat its motion in the x and y directions independently. In the x-direction, there is no acceleration, so the initial horizontal velocity will remain constant. In the y-direction, the ball will experience a downward acceleration due to gravity. Using the equations of projectile motion, we can find the time of flight and the vertical velocity of the baseball when it reaches the ground.
First, let's find the time of flight. We can use the equation:
h = v_y0 * t + (1/2) * g * t^2
where h is the initial height of the baseball, v_y0 is the initial vertical velocity, g is the acceleration due to gravity, and t is the time of flight. Substituting the given values, we have:
13.5 m = 13 m/s * sin(-21°) * t + (1/2) * 9.8 m/s^2 * t^2
Simplifying this equation and solving for t, we find that t ≈ 1.9 seconds.
Next, let's find the vertical velocity of the baseball when it reaches the ground. We can use the equation:
v_y = v_y0 + g * t
Substituting the given values, we have:
v_y = 13 m/s * sin(-21°) + 9.8 m/s^2 * 1.9 s
Simplifying this equation, we find that v_y ≈ -13.8 m/s. The negative sign indicates that the velocity is directed downward.
Finally, we can find the speed of the baseball when it reaches the ground by using the Pythagorean theorem:
speed = sqrt(v_x^2 + v_y^2)
Since the ball is thrown horizontally, the initial vertical velocity is 0, so v_x = 13 m/s * cos(-21°). Substituting the values, we have:
speed = sqrt((13 m/s * cos(-21°))^2 + (-13.8 m/s)^2)
Simplifying this equation, we find that the speed of the baseball when it reaches the ground is approximately 20.9 m/s.