A pitched ball is hit by a batter at a 50° angle and just clears the outfield fence, 97 m away. Assume that the fence is at the same height as the pitch and find the velocity of the ball when it left the bat. Neglect air resistance.
whats the answer
m/s at 50 degrees
t is time to top of parabola
u is constant horizontal speed
v is variable vertical speed
S is speed leaving the bat
then
u = S cos 50
top of arc is 97/2 from bat
so
97/2 = u t
and t = 97/(2 S cos 50)
vi = initial v = S sin 50
v = vi - g t = S sin 50 - 9.8 t
at top, 97/2 meters away, v = 0
0 = S sin 50 - 9.8 (97/(2 S cos 50))
solve for S
To find the velocity of the ball when it left the bat, we can use the equations of motion.
First, let's consider the vertical component of the ball's motion. Since the ball is clearing the fence, which is at the same height as the pitch, we know that the vertical displacement is zero.
Using the equation for vertical displacement:
Δy = v_iy * t + (1/2) * a * t^2
Since the vertical displacement is zero and we neglect air resistance, the acceleration in the vertical direction is equal to -9.8 m/s^2 (due to gravity).
0 = v_iy * t + (1/2) * (-9.8) * t^2
To solve for t, we can use the fact that the ball will hit the ground at t = 2 * t_total, where t_total is the total time of flight. Rearranging the equation:
t_total = 2t = 2 * (v_iy / 9.8)
Next, let's consider the horizontal component of the ball's motion. The horizontal displacement is given as 97 m, and the velocity in the horizontal direction remains constant throughout the motion.
Using the equation for horizontal displacement:
Δx = v_ix * t
Rearranging the equation:
v_ix = Δx / t_total = 97 m / (2 * t_total)
Finally, to find the initial velocity of the ball when it left the bat, we use the following trigonometric relation:
v_i = √(v_ix^2 + v_iy^2)
Substituting the values we found earlier:
v_i = √((97/(2*t_total))^2 + v_iy^2)
Now, let's find v_iy using the given angle of 50 degrees:
v_iy = v_i * sin(θ)
where θ is the angle of 50 degrees.
v_i = v_iy / sin(θ)
Now we can substitute this value back into the equation for v_i:
v_i = √((97/(2*t_total))^2 + (v_iy / sin(θ))^2)
Simplifying further:
v_i = √(97^2 / (4*t_total^2) + (v_iy^2 / sin(θ)^2))
Now we have the complete equation to find the velocity of the ball when it left the bat. We just need to substitute the values of θ, Δx, and solve for v_i using the equation:
v_i = √(97^2 / (4*t_total^2) + (v_iy^2 / sin(θ)^2))
To find the initial velocity of the ball when it left the bat, we can use the following equations of motion:
1. Horizontal motion equation: x = v₀ * t * cos(θ)
2. Vertical motion equation: y = v₀ * t * sin(θ) - (1/2) * g * t^2
Where:
- x is the horizontal distance traveled by the ball (97 m in this case),
- y is the vertical height achieved by the ball (equal to the height of the fence, assuming it is at the same height as the pitch),
- v₀ is the initial velocity of the ball,
- θ is the launch angle of the ball (50° in this case),
- g is the acceleration due to gravity (approximately 9.8 m/s²).
From the given information, we know that the ball just clears the outfield fence, so it reaches the height of the fence. This means that the vertical distance traveled by the ball is zero. Therefore, we can set y = 0 in the vertical motion equation.
Using this information, we can solve for v₀:
0 = v₀ * t * sin(θ) - (1/2) * g * t^2
We also know that the horizontal distance traveled by the ball is 97 m. Therefore, we can find the time, t, it took for the ball to reach that distance using the horizontal motion equation:
97 m = v₀ * t * cos(θ)
Now we have two equations and two unknowns (v₀ and t). Let's solve them simultaneously:
Equation 1: 0 = v₀ * t * sin(θ) - (1/2) * g * t^2
Equation 2: 97 m = v₀ * t * cos(θ)
Rearranging Equation 2, we get:
t = 97 m / (v₀ * cos(θ))
Now substitute this expression for t in Equation 1:
0 = v₀ * (97 m / (v₀ * cos(θ))) * sin(θ) - (1/2) * g * (97 m / (v₀ * cos(θ)))^2
Simplifying this equation will give us v₀, the initial velocity of the ball when it left the bat.