Two bodies are projected from ground with equal speeds 50ms from the s

1.	Two bodies are projected from ground with equal speeds 50ms from the same position in the same vertical plane tohave equal range but at different angles above the horizontal.If one of the angle is 30°, then the sum of their maximumheights is (g = 10 ms).
Answer» Answer: Given: 2 PROJECTILES are thrown with 50° at different angles such that they have same range. To find: Sum of max heights : Concept: At equal initial velocity, same range is possible only when the angles of projectiles are COMPLIMENTARY . So if ONE angle is 30°, then another angle will be 60°. Calculation: Max HEIGHT of first Projectiles : $h1 = \dfrac{ {u}^{2} {sin}^{2} ( \theta) }{2g} \\ = > h1 = \frac{ {50}^{2} \times { \sin(30 \degree) }^{2} }{20}$ $= > h1 = \dfrac{2500 \times \frac{1}{4} }{20} \\ = > h1 = \frac{125}{4} \\ = > h1 = 31.25 \: metres$ Now for 2nd Projectile : $h2 = \dfrac{ {u}^{2} {sin}^{2} ( \theta) }{2g} \\ = > h1 = \frac{ {50}^{2} \times { \sin(60 \degree) }^{2} }{20}$ $= > h1 = \dfrac{2500 \times \frac{3}{4} }{20} \\ = > h1 = \frac{125 \times 3}{4} \\ = > h1 = 93.75 \: metres$ So, total height = 31.25 + 93.75 => Total height = 125 metres. So final answer : $\boxed{ \blue{sum \: of \: heights = 125 \: m}}$

Two bodies are projected from ground with equal speeds 50ms from the same position in the same vertical plane tohave equal range but at different angles above the horizontal.If one of the angle is 30°, then the sum of their maximumheights is (g = 10 ms).

Answer»

2 PROJECTILES are thrown with 50° at different angles such that they have same range.

Sum of max heights :

At equal initial velocity, same range is possible only when the angles of projectiles are COMPLIMENTARY .

So if ONE angle is 30°,

then another angle will be 60°.

Max HEIGHT of first Projectiles :

$h1 = \dfrac{ {u}^{2} {sin}^{2} ( \theta) }{2g} \\ = > h1 = \frac{ {50}^{2} \times { \sin(30 \degree) }^{2} }{20}$

$= > h1 = \dfrac{2500 \times \frac{1}{4} }{20} \\ = > h1 = \frac{125}{4} \\ = > h1 = 31.25 \: metres$

Now for 2nd Projectile :

$h2 = \dfrac{ {u}^{2} {sin}^{2} ( \theta) }{2g} \\ = > h1 = \frac{ {50}^{2} \times { \sin(60 \degree) }^{2} }{20}$

$= > h1 = \dfrac{2500 \times \frac{3}{4} }{20} \\ = > h1 = \frac{125 \times 3}{4} \\ = > h1 = 93.75 \: metres$

So, total height = 31.25 + 93.75

=> Total height = 125 metres.

So final answer :

$\boxed{ \blue{sum \: of \: heights = 125 \: m}}$