Question

The potential energy of a projectile at its maximum height is equal to is kinetic energy there. If the velocity of projection is 20m/s, is time of fight is (g=10m/s²)

Answer 1

At maximum height, velocity of the particle = 20 cos a

Also given:

Potential energy of the particle at maximum height = kinetic energy

Which is, $m\times g\times h=\frac { 1 }{ 2 } m\times v^{ 2 }$

$10\times h=\frac { 1 }{ 2 } \times (20 cosa)^{ 2 }$

$h=\frac { (1\times 400cos^{ 2 }a) }{ 2\times 10 } =20cos^{ 2 }a\quad\rightarrow (1)$

Height of the projectile $h=\frac { [v^{ 2 }sin^{ 2 }a] }{ 2 } g\quad\rightarrow (2)$

Substitute (1) in (2)

$20cos^{ 2 }a=\frac { 20^{ 2 }sin^{ 2 }a }{ 2\times (10) }$

$cos^{ 2 }a=\frac { 20sin^{ 2 }a }{ 20 }$

$cos^2 a = sin^2 a$

$\frac { sin^{ 2 }a }{ cos^{ 2 }a } =1$

$tan^2 a = 1$

tan a = 1, a = 45°

$Time=\frac { (2\times u sina) }{ g } =\frac { (2\times 20\times sin45) }{ 10 } =\frac { 4\times 1 }{ \sqrt { 2 } } =2\sqrt { 2 } seconds$

Answer 2

Answer:

*your answer is "2√2" *

Explanation:

GIVEN ,

• P.E=K.E

• 1/2mu²sin²θ=1/2mu²cos²θ

• sin²θ=cos²θ

• θ=45°

• t≈2u.sinθ/g
• t=2.20.sin45/10
• t=20√2/10
• t=2√2

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