Question Statement

A batsman hits the ball. The ball attains the maximum height of $25\mathrm{m}$ and drops on the ground at a distance of $80\mathrm{m}$ from the batsman. Assuming the origin at the position of batsman:

1. Write the equation of path of the ball.
2. Is it possible for a man of height $1.6\mathrm{m}$ to catch the ball, standing $70\mathrm{m}$ away from the batsman?

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Background and Explanation

This problem involves projectile motion, where the trajectory of a projectile (neglecting air resistance) follows a parabolic path. The vertex of the parabola represents the maximum height, and the ball lands on the ground at the range of the projectile.

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Solution

Part 1: Equation of the Path

The path of the ball is parabolic since it is a projectile. According to given conditions, the coordinates of the vertex $V$ (maximum height point) are $(40,25)$, since the maximum height occurs at half the horizontal range.

Let the equation of the parabola with vertex $(h,k)=(40,25)$ be:

$(x-40{)}^2=-4a(y-25)$

Since the point $A(80,0)$ lies on the parabola (where the ball hits the ground), we substitute $x=80$ and $y=0$:

$\begin{array}{rl}(80-40{)}^2& =-4a(0-25)\\ (40{)}^2& =100a\\ 1600& =100a\\ a& =16\end{array}$

Substituting $a=16$ back into equation (i):

$\begin{array}{rl}(x-40{)}^2& =-4(16)(y-25)\\ (x-40{)}^2& =-64(y-25)\end{array}$

This is the required equation of the path of the ball.

Part 2: Possibility of Catching the Ball

To determine if a man standing $70\mathrm{m}$ away from the batsman can catch the ball, we calculate the height of the ball at $x=70$.

Substituting $x=70$ into the path equation:

$\begin{array}{rl}(70-40{)}^2& =-64(y-25)\\ 30^2& =-64y+1600\\ 900& =-64y+1600\\ 64y& =1600-900\\ 64y& =700\\ y& =\frac{700}{64}=10.93\mathrm{m}\end{array}$

Thus, at a distance of $70\mathrm{m}$ from the batsman, the height of the ball is $10.93\mathrm{m}$. Since $10.93\mathrm{m}>1.6\mathrm{m}$ (the man's height), the man cannot catch the ball.

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Key Formulas or Methods Used

• Standard equation of downward parabola with vertex $(h,k)$: $(x-h{)}^2=-4a(y-k)$
• Vertex coordinates for symmetric projectile: $\left(\frac{R}{2},H_{\max }\right)$ where $R$ is the range and $H_{\max }$ is the maximum height
• Substitution method: Finding unknown parameters using known points on the curve

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Summary of Steps

1. Identify the vertex: The maximum height of $25\mathrm{m}$ occurs at the midpoint of the $80\mathrm{m}$ range, so $V=(40,25)$.
2. Set up the parabola equation: Use $(x-40{)}^2=-4a(y-25)$.
3. Find parameter $a$: Substitute the landing point $(80,0)$ to solve $1600=100a$, giving $a=16$.
4. Write final equation: $(x-40{)}^2=-64(y-25)$.
5. Check height at $x=70$: Substitute to find $y=\frac{900}{64}\approx 10.93\mathrm{m}$.
6. Compare heights: Since $10.93\mathrm{m}>1.6\mathrm{m}$, the man cannot catch the ball.