Question

a body projected horizontally from the top of the tower follows =20x^2 parabola equation where x,y, are in m (g=10m/s^2)then the velocity of the projectile is​

Answer 1

By horizontal motion of the projectile, we have,

x = u cosθ t

Then,

t = x / u cosθ

By vertical motion of the projectile, we have,

y = u sinθ t - (g t²) / 2

y = u sinθ (x / u cosθ) - (g x² / u² cos²θ) / 2

y = x tanθ - (g / (2 u² cos²θ)) x²

This is the normal equation for the trajectory.

If the body is projected horizontally, θ = 0°. Then the equation will be,

y = g x² / (2 u²) → (1)

where the acceleration is g, not - g.

In the question, the equation of the trajectory is,

y = 20 x²

Comparing this with (1), we get that,

g / (2 u²) = 20

Taking g = 10 m s^(-2),

u² = 10 / (2 × 20)

u² = 1 / 4

u = 0.5 m s^(-1)

Answer 2

Answer:

By horizontal motion of the projectile, we have,

x = u cosθ t

Then,

t = x / u cosθ

By vertical motion of the projectile, we have,

y = u sinθ t - (g t²) / 2

y = u sinθ (x / u cosθ) - (g x² / u² cos²θ) / 2

y = x tanθ - (g / (2 u² cos²θ)) x²

This is the normal equation for the trajectory.

If the body is projected horizontally, θ = 0°. Then the equation will be,

y = g x² / (2 u²) → (1)

where the acceleration is g, not - g.

In the question, the equation of the trajectory is,

y = 20 x²

Comparing this with (1), we get that,

g / (2 u²) = 20

Taking g = 10 m s^(-2),

u² = 10 / (2 × 20)

u² = 1 / 4

u = 0.5 m s^(-1)

Explanation: