Question

Two stones thrown at different angles have same initial velocity and same range. If H is the maximum height attained by one stone thrown at an angle of 30°, then the maximum height attained by the other stone is (b) H (c) 2H (d) 3H

Answer 1

Explanation:

solution is attached herewith

Answer 2

MAX HEIGHT OF 2ND STONE IS 3H (option d)

Given:

• Both stones have same range.
• Angle of projection of the first stone is 30°
• Max height reached by the stone is H.

To find:

• Height reached by the other stone will be?

Calculation:

We know that :

• For a constant value of velocity, two projectiles will have same range for complementary angle of projection.

For 1st stone :

$H = \dfrac{ {u}^{2} { \sin}^{2} ( {30}^{ \circ} )}{2g}$

$\implies H = \dfrac{ {u}^{2} \times \dfrac{1}{ 4 } }{2g}$

$\implies H = \dfrac{ {u}^{2} }{8g}$

• Now, the second stone the angle of projection will be 90°-30° = 60°.

So, max height will be :

$H' = \dfrac{ {u}^{2} { \sin}^{2} ( {60}^{ \circ} )}{2g}$

$\implies H '= \dfrac{ {u}^{2} \times \dfrac{3}{ 4 } }{2g}$

$\implies H '= \dfrac{3 {u}^{2} }{8g}$

$\implies H '=3 \times \dfrac{ {u}^{2} }{8g}$

$\implies H '=3 H$

So, maximum height attained by the second stone is 3H.