An outfielder throws a 0.145 kg baseball to the shortstop with a speed of 30 m/s at an angle of 20 degrees above the horizontal. How far does the ball travel? (Assume the shortstop catches it
at about the same height that the outfielder throws it.)
nah
To find the distance the ball travels, we need to break down its initial velocity into horizontal and vertical components. The horizontal component represents the distance traveled along the x-axis, and the vertical component represents the distance traveled along the y-axis.
First, let's calculate the initial horizontal velocity (Vx) and the initial vertical velocity (Vy) using the given angle and speed.
Vx = V * cos(θ)
Vy = V * sin(θ)
where:
V is the magnitude of the initial velocity (30 m/s in this case)
θ is the angle of projection (20 degrees in this case)
Plugging in the values, we get:
Vx = 30 m/s * cos(20°)
Vy = 30 m/s * sin(20°)
Next, we need to find the time of flight (t) of the ball. Since the ball is caught at the same height it is thrown, the time taken to reach the shortstop is the same as the time taken to reach the maximum height.
Using the equation:
t = 2 * Vy / g
where:
g is the acceleration due to gravity (approximately 9.8 m/s^2), we get:
t = 2 * (30 m/s * sin(20°)) / 9.8 m/s^2
Finally, we can calculate the horizontal distance (D) traveled by the ball:
D = Vx * t
Plugging in the values we've calculated, we can solve for D:
D = (30 m/s * cos(20°)) * (2 * (30 m/s * sin(20°)) / 9.8 m/s^2)
Evaluating this expression gives us the distance traveled by the ball.