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

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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Orlando Paz
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
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.
user_27qwe

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 $$\cos^{2}45^{\circ}=\frac{1}{2}$$
$$clearance=d-\frac{d^{2}}{R}-h_{fence}+h_{bat}=97.5m-\frac{(97.5m)^{2}}{107m}-7.32m+1.22m\approx2.56m$$