A batter hits a pitched ball when the center of the ball is 1.22 m abo

Priscilla Johnston 2021-12-26 Answered
A batter hits a pitched ball when the center of the ball is 1.22 m above the ground. The ball leaves the bat at an angle of \(\displaystyle{45}^{{\circ}}\) with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 107 m.
a) Does the ball clear a 7.32-m-high fence that is 97.5 m horizontally from the launch point?
b) At the fence, what is the distance between the fence top and the ball center?

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Expert Answer

Orlando Paz
Answered 2021-12-27 Author has 716 answers
Step 1
Givens:
The Distance from the origin to start point is: \(\displaystyle{y}_{{{0}}}={1.22}{m}\)
The launch angle is: \(\displaystyle\theta_{{{0}}}={45}^{{\circ}}\)
The Horizontal rangle is: \(\displaystyle{R}={107}{m}\)
Step 2
Part a
For 7.32 m high fence, the Horizontal distance from the origin is:
\(\displaystyle{x}-{x}_{{{0}}}={97.5}{m}\)
To determine the clearance over the high fence, we use
\(\displaystyle\Delta{y}={y}-{h}\)
So firstly, we need to find the maximum height:
\(\displaystyle{y}-{y}_{{{0}}}={\left({\tan{\theta}}_{{{0}}}\right)}{x}-{\frac{{{g}{x}^{{{2}}}}}{{{2}{\left({v}_{{{0}}}{\cos{\theta}}_{{{0}}}\right)}^{{{2}}}}}}\)
We will use the Horizontal rangle:
\(\displaystyle{R}={\frac{{{{v}_{{{0}}}^{{{2}}}}}}{{{g}}}}{\sin{{2}}}\theta_{{{0}}}\)
\(\displaystyle{107}{m}={\frac{{{{v}_{{{0}}}^{{{2}}}}}}{{{9.8}{\frac{{{m}}}{{{s}^{{{2}}}}}}}}}{\sin{{\left({2}\times{45}^{{\circ}}\right)}}}\)
\(\displaystyle{v}_{{{0}}}={32.4}{\frac{{{m}}}{{{s}}}}\)
Step 3
Then,
\(\displaystyle{y}={1.22}{m}+{\left({\tan{{45}}}^{{\circ}}\right)}{\left({97.5}\right)}-{\frac{{{\left({9.8}{\frac{{{m}}}{{{s}^{{{2}}}}}}\right)}{\left({97.5}{m}\right)}^{{{2}}}}}{{{2}{\left({32.4}{\frac{{{m}}}{{{s}}}}{\cos{{45}}}^{{\circ}}\right)}^{{{2}}}}}}\)
\(\displaystyle={10}{m}\)
The clearance of the ball can be obtained from:
\(\displaystyle\Delta{y}={y}-{h}\)
\(\displaystyle={10}{m}-{7.32}{m}\)
\(\displaystyle={2.7}{m}\)
Yes, the ball clear above the fence.
Step 4
Part b:
The center of the ball at: 10 m
The height of the fence is: 7.32 m
Then, the distance between the ball center and the fence top is: (clearance of the ball)
\(\displaystyle\Delta{y}={y}-{h}\)
\(\displaystyle={10}{m}-{7.32}{m}\)
\(\displaystyle\Delta{y}={2.7}{m}\)
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William Appel
Answered 2021-12-28 Author has 5636 answers
Step 1
The formula for the range of a projectile is:
\(\displaystyle{R}={\frac{{{{v}_{{{0}}}^{{{2}}}}}}{{{g}}}}{\sin{{2}}}\theta\)
Substitute the known values and determine the initial velocity.
\(\displaystyle{107}{m}={\frac{{{{v}_{{{0}}}^{{{2}}}}}}{{{\left({9.81}{\frac{{{m}}}{{{s}^{{{2}}}}}}\right)}}}}{\sin{{2}}}{\left({45}^{{\circ}}\right)}\)
\(\displaystyle{v}_{{{0}}}=\sqrt{{{\frac{{{\left({107}{m}\right)}{\left({9.81}{\frac{{{m}}}{{{s}^{{{2}}}}}}\right)}}}{{{\sin{{2}}}{\left({45}^{{\circ}}\right)}}}}}}\)
\(\displaystyle={32.4}{\frac{{{m}}}{{{s}}}}\)
Step 2
The time required to cover the 97.5 m Horizontal distance is:
\(\displaystyle{t}={\frac{{{x}}}{{{v}_{{{x}}}}}}\)
\(\displaystyle={\frac{{{x}}}{{{v}_{{{0}}}{\cos{{45}}}^{{\circ}}}}}\)
\(\displaystyle={\frac{{{97.5}{m}}}{{{\left({32.4}{\frac{{{m}}}{{{s}}}}\right)}{\left({\frac{{{1}}}{{\sqrt{{{2}}}}}}\right)}}}}\)
\(\displaystyle={4.256}{s}\)
Step 3
The vertical distance covered by the ball in till the time it covers 97.5 m
Horizontal distance is:
\(\displaystyle{y}-{y}_{{{0}}}={v}_{{{0}{y}}}{t}-{\frac{{{1}}}{{{2}}}}{>}^{{{2}}}\)
\(\displaystyle{y}={1.22}+{\left({32.4}\frac{{m}}{{s}}\right)}{\left({\sin{{45}}}^{{\circ}}\right)}{\left({4.256}{s}\right)}-{\frac{{{1}}}{{{2}}}}{\left({9.81}\frac{{m}}{{s}^{{{2}}}}\right)}{\left({4.256}{s}\right)}^{{{2}}}\)
\(\displaystyle={8.7}{m}\)
\(\displaystyle={10}{m}\)
Thus, the ball will clear the 7.32 m high fence.
0
user_27qwe
Answered 2021-12-30 Author has 9558 answers

Step 1
Find: Whether a batted baseball clears a fence, and by what amount it does or does not.
Given: The baseball’s initial launch height and angle, the range the baseball would have without the fence, the distance to the fence and its height.
Let the y axis run vertically and the x axis horizontally. Let the range the baseball would have without the fence be \(R=107 m\), with the distance to the fence \(d=97.5m\) and its height \(h_{fence}= 7.32 m\). The baseball is batted at an angle \(\theta=45^{\circ}\) at speed vi a height of \(h_{bat}= 1.22m\) above the ground.
Let the origin be at the position the ball leaves the bat. The height of the fence relative to the height of the bat is then
Step 2
\(\delta h=h_{fence}-h_{bat}\)
What we really need to determine is the ball’s y coordinate at \(x = d\). If \(y > \delta \), the ball clears the fence. We can use the range the baseball would have without the fence and the launch angle to find the ball’s speed, which will allow a complete calculation of the trajectory.
Relevant equations: We need only the equations for the range and trajectory of a projectile over level ground:
\(R=\frac{v_{i}^{2}\sin2\theta}{g}\)
For convenience sake and easy reading, I extracted the solution of your textbook for the remaining parts of the solution.
Hence, it is seen that the ball does clear the fence, by approximately 2.56 m
Step 3
\(R=\frac{v_{i}^{2}\sin2\theta}{g}\)
\(y(x)=x\tan\theta-\frac{gx^{2}}{2v_{i}^{2}\cos^{2}\theta}\)
Symbolic solution: From the range equation above, we can write the velocity in terms of known quantities:
\(v_{i}=\sqrt{\frac{R_{g}}{\sin2\theta}}\)
The trajectory then becomes \(y(x)=x\tan\theta-\frac{gx^{2}\sin2\theta}{2Rg\cos^{2}\theta}=x\tan\theta-\frac{x^{2}\sin2\theta}{2R\cos^{2}\theta}\)
The height difference between the ball and the fence is \(y(d)-\delta h\). If it is positive, the ball clears the fence.
\(clearence=y(d)-\delta h=d\tan\theta-\frac{d^{2}\sin2\theta}{2R\cos^{2}\theta}-\delta h\)
\(=d\tan\theta-\frac{d^{2}\sin2\theta}{2R\cos^{2}\theta}-h_{fence}+h_{bat}\)
Numeric solution: Using the numbers given, and noting:
\(\tan 45^{\circ}=1,\sin90^{\circ}=1\) and \(\cos^{2}45^{\circ}=\frac{1}{2}\)
\(clearance=d-\frac{d^{2}}{R}-h_{fence}+h_{bat}=97.5m-\frac{(97.5m)^{2}}{107m}-7.32m+1.22m\approx2.56m\)

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