The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance up to which he can throw the same ball is:

Given: Maximum vertical height is H max = 136 m. Find: The maximum horizontal range for the same ball. For a vertical throw, the maximum height is H max = v^22g So, v^2 = 2gH max The maximum horizontal range for the same speed is R max = v^2g Substituting v^2 = 2gH max, R max = 2gH maxg = 2H max Therefore, R max = 2 136 = 272 m Therefore, the maximum horizontal distance is 272 m and the correct option is C. The solution labels the correct option as D, but its working gives 272 m, which matches option C.

The maximum vertical height to which a man can throw a ball is 13… — Solution | JEE Physics

MCQEasyJEE 2023Projectile Motion

The maximum vertical height to which a man can throw a ball is $136 \, \text{m}$ . The maximum horizontal distance up to which he can throw the same ball is:

A
$192 \, \text{m}$
B
$136 \, \text{m}$
C
$272 \, \text{m}$
D
$68 \, \text{m}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Maximum vertical height is $H_{\text{max}} = 136 \, \text{m}$ .

Find: The maximum horizontal range for the same ball.

For a vertical throw, the maximum height is

H_{\text{max}} = \frac{v^2}{2g}

So,

v^2 = 2gH_{\text{max}}

The maximum horizontal range for the same speed is

R_{\text{max}} = \frac{v^2}{g}

Substituting $v^2 = 2gH_{\text{max}}$ ,

R_{\text{max}} = \frac{2gH_{\text{max}}}{g} = 2H_{\text{max}}

Therefore,

R_{\text{max}} = 2 \times 136 = 272 \, \text{m}

Therefore, the maximum horizontal distance is $272 \, \text{m}$ and the correct option is C.

The solution labels the correct option as D, but its working gives $272 \, \text{m}$ , which matches option C.

Common mistakes

Using the formula for range at an arbitrary angle instead of the maximum range. Maximum range occurs at $45^\circ$ , so for the same speed one must use $R_{\text{max}} = \frac{v^2}{g}$ .
Forgetting that the given $136 \, \text{m}$ is the maximum vertical height for a vertical throw, so it must be related through $H_{\text{max}} = \frac{v^2}{2g}$ . Treating it directly as range gives the wrong result.
Missing the relation $R_{\text{max}} = 2H_{\text{max}}$ for the same projection speed. Once $H_{\text{max}}$ is known, the range is not equal to it; it is twice that value.

Practice more Projectile Motion questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Projectile Motion

Answer

Step-by-step solution

Common mistakes

Practice more Projectile Motion questions

Related questions