A cricket player throws the ball to have the maximum horizontal range of 120m. If the throws the ball vertically with same velocity what is the maximum height it can reach ?

Answer

To solve the problem step by step, we will use the principles of projectile motion.

### Step 1: Understand the relationship between range and initial velocity
The maximum horizontal range \( R_{\text{max}} \) of a projectile is given by the formula:
\[ R_{\text{max}} = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection. For maximum range, \( \theta \) should be \( 45^\circ \), which makes \( \sin(90^\circ) = 1 \).

### Step 2: Set up the equation for maximum range
Since the maximum range is given as \( 120 \, \text{m} \):
\[ R_{\text{max}} = \frac{u^2}{g} \]
Substituting the known values:
\[ 120 = \frac{u^2}{10} \]

### Step 3: Solve for initial velocity \( u \)
Rearranging the equation gives:
\[ u^2 = 1200 \]
Taking the square root:
\[ u = \sqrt{1200} = 34.64 \, \text{m/s} \, (\text{approximately}) \]

### Step 4: Find the maximum height when thrown vertically
The maximum height \( h_{\text{max}} \) reached by a projectile is given by the formula:
\[ h_{\text{max}} = \frac{u^2}{2g} \]
Substituting the value of \( u^2 \):
\[ h_{\text{max}} = \frac{1200}{2 \times 10} \]
Calculating this gives:
\[ h_{\text{max}} = \frac{1200}{20} = 60 \, \text{m} \]

### Final Answer
The maximum height the ball can reach when thrown vertically with the same velocity is \( 60 \, \text{m} \).

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