Question

Q Prove that horizontal range of an oblique projectile fired at an angle Θ and ( 90-Θ) is same, hence obtain expression for maximum height, also show that sum of maximum heights for these two angles is independent of angle of projection.

11th physics please help me

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Answer 1

$\huge\bigstar\underline\pink{Answer}$

★ For angle ∅ Range =

R = V°²Sin 2∅ / g

★ For angle (90-∅)

R = V°²Sin2(90-∅)/g

R = V°² Sin (180-∅)/g

$\pink{sin180-∅\:=\:sin∅}$

R = V°²Sin2∅/g

hence, range is same for both angles of projection ∅ and (90-∅)

★ The range will be same if sum of angle of projection is 90°

★ if Question asks for two different projections , we use formula

Hmax = Rmax /2

As range is same , so maximum height will be same. ✓✓

★Expression for maximum height

→ From Newton's law of motion

V² = U² + 2as

here,

U = 0

V= V° sin ∅

a = g

we, get H = V°² sin²∅ /2g

maxm value of sin = sin90°= 1

so, maximum height =

H = V²/2g

Happy to help!

☺️

Answer 2

Answer:

★ For angle ∅ Range =

R = V°²Sin 2∅ / g

★ For angle (90-∅)

R = V°²Sin2(90-∅)/g

R = V°² Sin (180-∅)/g

\pink{sin180-∅\:=\:sin∅}

R = V°²Sin2∅/g

hence, range is same for both angles of projection ∅ and (90-∅)

★ The range will be same if sum of angle of projection is 90°

★ if Question asks for two different projections , we use formula

Hmax = Rmax /2

As range is same , so maximum height will be same. ✓✓

★Expression for maximum height

→ From Newton's law of motion

V² = U² + 2as

here,

U = 0

V= V° sin ∅

a = g

we, get H = V°² sin²∅ /2g

maxm value of sin = sin90°= 1

so, maximum height =

H = V²/2g