A baseball is hit at ground level. The ball is observed to reach its maximum height above ground level 3.0 s after being hit. And 2.5 s after reaching this maximum height, the ball is observed to barely clear a fence that is 97.5 m from where it was hit. How high is the fence?
First, you want to find the maximum height.
At the maximum height, you know the velocity must be 0.
vi = initial velocity
vy(t) = vertical velocity
t1 = first time (3 s)
SO:
vy(t1) = -g*t1 + vi*sinθ
0 = -g*t1 + vi*sinθ
vi*sinθ = g*t1
Now, it's just a matter of substituting that into the equation for vertical distance.
ymax = maximum height reached by ball
y(t) = vertical distance
ymax = y(t1) = -1/2*g*t1^2 + vi*sinθ*t1
= -1/2*g*t1^2 + g*t1*t1
= -1/2*g*t1^2 + g*t1^2
= 1/2*g*t1^2
= 1/2(9.8)3^2 = 44.1m
From there, it's a simple matter of finding vi*sinθ.
44.1 = -1/2*g*t1^2 + vi*sinθ*t1
44.1 = -1/2(9.8)3^2 + vi*sinθ(3)
vi*sinθ = 29.4
When t = 5.5
y(5.5) = -1/2(9.8)5^2 + 29.4(5.5)
= 13.475m
Thus, in order for the ball to clear the fence, it must be less than 13.475 m!
Horizontal:
97.5=vi*5.5
so vihorizontal= 97.5/5.5 m/s
vertical: at the top, the vertical veloicty is zero, so
0=viy-1/2 g*3^3
viy=4.9*9g
so now we have the vertical and horizontla components.
finally
hf=viy^t-1/2 g t^2 so at t=5.5, knowing viy, calculate hf.
5.5
To find the height of the fence, we need to analyze the motion of the baseball.
Let's break down the problem into two parts: the upward motion of the baseball and the downward motion of the baseball.
1. Upward Motion:
The time it takes for the ball to reach its maximum height is 3.0 seconds. At this point, the ball is momentarily at rest before it starts coming back down. Since the ball is only traveling in the vertical direction, we can use the basic kinematic equation:
h1 = v0 * t + (1/2) * a * t^2
Where:
h1 is the maximum height,
v0 is the initial vertical velocity,
t is the time,
a is the vertical acceleration (which is -9.8 m/s^2 due to gravity).
Since the ball starts at ground level, the initial height h0 is 0, and the initial vertical velocity v0 is unknown. We need to find v0 to solve this part of the equation.
We will use the fact that the total time to reach the maximum height is 3.0 seconds and divide it into two equal parts since the upward and downward times are symmetrical. Therefore, the time to reach the maximum height is 1.5 seconds.
So the equation becomes:
h1 = v0 * 1.5 + (1/2) * (-9.8) * (1.5)^2
2. Downward Motion:
After reaching the maximum height, the ball starts coming back down. The remaining time to clear the fence is 2.5 seconds. The equation to find the distance covered in the downward motion is:
h2 = v0 * t + (1/2) * a * t^2
Where:
h2 is the height of the fence,
v0 is the initial vertical velocity,
t is the time (2.5 seconds), and
a is the vertical acceleration (-9.8 m/s^2).
We need to find h2 to determine the height of the fence.
Now, we can solve these two equations simultaneously to find both v0 and h2:
Equation 1: h1 = v0 * 1.5 + (1/2) * (-9.8) * (1.5)^2
Equation 2: h2 = v0 * 2.5 + (1/2) * (-9.8) * (2.5)^2
By substituting the value of h1 from Equation 1 into Equation 2, we can solve for h2.
Solving these equations will give you the height of the fence.