A pitched ball is hit by a batter at a 40° angle and just clears the outfield fence, 103 m away. If the fence is at the same height as the pitch, find the velocity of the ball when it left the bat. Ignore air resistance.
recall that the range of such a projectile is
R = v^2/g sin2θ
so plug in your numbers and solve for v
Initial speed up = Vi = S sin 40 = 0.643 S
initial (and final) horizontal speed = u = S cos 40 = 0.766 S
Horizontal problem:
103 = 0.766 S t where t is time in air
Vertical problem:
h = 0 = Vi t - (1/2) (9.81) t^2
0 =0 0.643 S t - 4.9 t^2 = t ( .643 S - 4.9t )
well we all know the height is 0 when t is 0 so we need
.643 S = 4.9 t
but we know 103 = 0.766 S t from the horizontal problem so
.643 S = 4.9 [ 103 / ( 0.766S) ]
0.493 S^2 = 505
S = 32 meters / second at 40 deg above horizontal
To find the velocity of the ball when it left the bat, we can use the equations of projectile motion. Projectile motion involves two independent motions: horizontal motion (along the x-axis) and vertical motion (along the y-axis).
First, let's break down the given information:
- Angle of projection (θ) = 40°
- Distance traveled (horizontal displacement, x) = 103 m
- Height of the fence (vertical displacement, y) = 0 m (same height as the pitch)
Since there is no air resistance, the time taken for the ball to reach the fence in the horizontal direction is the same as the time taken for the ball to reach the maximum height in the vertical direction.
Let's find the initial velocity of the ball (Vo) by considering the vertical motion:
- The vertical velocity component (Vy) of the ball at its maximum height is 0 because it comes to a stop momentarily.
- The acceleration due to gravity (g) is approximately 9.8 m/s².
Using the equation Vy = Vo * sin(θ) - g * t, where t represents the time of flight, we can determine the time taken for the ball to reach its maximum height.
Since Vy = 0 at its maximum height, the equation becomes 0 = Vo * sin(θ) - g * t.
Rearranging the equation to solve for t, we get t = Vo * sin(θ) / g.
Now, let's find the horizontal component of the initial velocity (Vox) using the equation x = Vox * t, where x represents the horizontal displacement and t is the time of flight.
Substituting the values into the equation, we get 103 = Vox * (Vo * sin(θ) / g).
Simplifying the equation, we have 103 = (Vox² * sin(2θ)) / g.
To separate Vox from the equation, we need to find the value of sin(2θ).
Using the double angle formula, sin(2θ) = 2 * sin(θ) * cos(θ).
Now, we can rewrite the equation as 103 = (Vox² * 2 * sin(θ) * cos(θ)) / g.
Since cos(θ) = √(1 - sin²(θ)), we can further simplify the equation to 103 = (Vox² * 2 * sin(θ) * √(1 - sin²(θ))) / g.
At this point, we have a trigonometric equation that involves sin(θ), which we know is equal to sin(40°).
Plugging in the values, we have 103 = (Vox² * 2 * sin(40°) * √(1 - sin²(40°))) / g.
We can now solve for Vox² by rearranging the equation as Vox² = (103 * g) / (2 * sin(40°) * √(1 - sin²(40°))).
Finally, to find the initial velocity of the ball (Vo), we take the square root of Vox², Vo = √(Vox²).
By substituting the known values of g = 9.8 m/s² and sin(40°), we can plug them into the equation and calculate the value of Vo.