A home run just clears a fence 105m from home plate. The fence is 4.00m higher than the height at which the batter struck the ball, and the ball left the bat at a 31.0 degree angle above the horizontal. At what speed did the ball leave the bat?
To determine the speed at which the ball left the bat, we need to use the principles of projectile motion. We will break down the problem into horizontal and vertical components and then combine them to calculate the initial speed.
First, let's determine the vertical component of the ball's velocity.
Given:
- The ball left the bat at a 31.0 degree angle above the horizontal.
- The fence is 4.00m higher than the height at which the batter struck the ball.
We need to find:
- The vertical component of the ball's velocity.
Using trigonometry, we can find the vertical component using the equation:
V_vertical = V * sin(θ)
where:
- V_vertical is the vertical component of the velocity.
- V is the initial velocity (unknown).
- θ is the angle given (31.0 degrees).
Now, let's calculate the vertical component:
V_vertical = V * sin(31.0 degrees)
Next, let's determine the time it takes for the ball to reach the fence.
We can use the following equations:
y = V_initial * t + (1/2) * g * t^2
where:
- y is the vertical distance the ball traveled (4.00m).
- g is the acceleration due to gravity (9.8 m/s^2).
- V_initial is the initial velocity in the vertical direction (V_vertical).
- t is the time.
In this case, we are solving for t.
Rearranging the equation, we have:
4.00 = V_vertical * t + (1/2) * g * t^2
Now, let's solve for t.
Using the quadratic formula, we have:
t = (-V_vertical ± sqrt(V_vertical^2 - 4*(1/2)*g*(-y))) / (2*(1/2)*g)
Since we are dealing with time, we can ignore the negative value.
Now that we have the time (t), let's calculate the horizontal distance traveled by the ball.
Using the equation:
x = V_horizontal * t
where:
- x is the horizontal distance traveled.
- V_horizontal is the horizontal component of the initial velocity.
- t is the time.
We need to find the horizontal component of the initial velocity (V_horizontal). We can use the equation:
V_horizontal = V * cos(θ)
where:
- V_horizontal is the horizontal component of the velocity.
- V is the initial velocity (unknown).
- θ is the angle given (31.0 degrees).
Using the equation above, let's calculate V_horizontal.
V_horizontal = V * cos(31.0 degrees)
Now, let's calculate x using the equation:
x = V_horizontal * t
Finally, we need to solve for the initial velocity (V).
We know that the horizontal distance traveled (x) is equal to 105 m. Thus, we have:
105 m = V * cos(31.0 degrees) * t
Now substitute the value of t from the previous equation and solve for V.
Substituting the value of t we found earlier, we have:
105 m = V * cos(31.0 degrees) * ((-V_vertical + sqrt(V_vertical^2 - 4*(1/2)*g*(-y))) / (2*(1/2)*g))
Now, we can simplify and solve this equation for V (the initial velocity).