A batter hits a baseball at an angle of 30 degrees with the horizontal. The ball is hit at a height of 2 m and leaves the bat with a speed of Vo. What is the minimum Vo for the ball to clear the outfield fence? The fence is 5 m high and located 60 m from home plate.
To find the minimum Vo required for the ball to clear the outfield fence, we can break down the problem into two components: the vertical and horizontal motion of the ball.
First, let's consider the vertical motion. The ball is launched at an angle of 30 degrees with the horizontal, and it needs to clear a fence that is 5 m high. We can use the equations of motion to find the time it takes for the ball to reach its maximum height and then come back down.
1. Vertical motion:
- Initial vertical velocity (Vy0) = Vo * sin(angle)
- Time to reach maximum height (t_max) = Vy0 / g
- Maximum height (h_max) = (Vy0^2) / (2 * g)
- Time of flight (t_total) = 2 * t_max
Now let's consider the horizontal motion. The ball needs to travel a horizontal distance of 60 m from home plate to clear the fence.
2. Horizontal motion:
- Horizontal distance (d) = Vo * cos(angle) * t_total
To clear the fence, the maximum height of the ball (h_max) needs to be greater than the height of the fence (5 m). This means:
h_max > 5 m
From equation 1, we can substitute the expression for t_max into the equation for h_max:
[ (Vo * sin(angle))^2 ] / (2 * g) > 5 m
Now we can solve for the minimum Vo:
Vo > sqrt( 5 * 2 * g / sin(angle)^2 )
Substituting the values of g = 9.8 m/s^2 and angle = 30 degrees, we can calculate the minimum Vo required for the ball to clear the fence.
To find the minimum initial velocity (Vo) required for the ball to clear the outfield fence, we can use the kinematic equations of motion.
We know the following information:
- Angle of launch (θ): 30 degrees
- Initial height (h): 2 m
- Height of the fence (H): 5 m
- Distance to the outfield fence (x): 60 m
We need to calculate the minimum initial velocity (Vo) required.
Step 1: Calculate the time of flight (t) of the ball.
Using the equation:
h = (Vo^2 * sin^2(θ)) / (2 * g)
where g is the acceleration due to gravity (9.8 m/s^2)
Rearranging the equation to solve for the time of flight (t):
t = sqrt((2 * h) / g)
Substituting the given values:
t = sqrt((2 * 2) / 9.8)
t ≈ 0.64 seconds
Step 2: Calculate the horizontal distance covered (R) by the ball.
Using the equation:
R = Vo * cos(θ) * t
Substituting the given values and the calculated time (t):
R = Vo * cos(30) * 0.64
R = 0.55Vo
Step 3: Calculate the height (Y) reached by the ball at the fence distance (x).
Using the equation:
Y = h + (Vo * sin(θ) * t) - (0.5 * g * t^2)
Substituting the given values and the calculated time (t):
Y = 2 + (Vo * sin(30) * 0.64) - (0.5 * 9.8 * 0.64^2)
Y = 2 + 0.32Vo - 1.26
Y = 0.32Vo + 0.74
Step 4: Determine the minimum Vo to clear the fence.
For the ball to clear the fence, its height at the distance (x) should be greater than the fence height (H).
Therefore, Y > H.
Substituting the given values:
0.32Vo + 0.74 > 5
Solving for Vo:
0.32Vo > 5 - 0.74
0.32Vo > 4.26
Vo > 4.26 / 0.32
Vo > 13.3125
Therefore, the minimum initial velocity (Vo) required for the ball to clear the outfield fence is approximately 13.3125 m/s.