Question

A man standing on a hill top projects a stone horizontally with speed v0 as shown in figure. Taking the coordinate system as given in the figure. The coordinates of the point where the stone will hit the hill surface​

Answer 1

Answer:

(2V₀²Tanθ/g , - 2V₀²Tan²θ/g)

Explanation:

A man standing on a hill top projects a stone horizontally with speed v0 as shown in figure. Taking the coordinate system as given in the figure. The coordinates of the point where the stone will hit the hill surface​

Horizontal speed = V₀

Vertical Speed = 0

Let say coordinate where stone hit the surface = ( x , -y)

S = ut + (1/2)at²

y =  (1/2)gt²

t² = 2y/g

t = √(2y/g)

Horizontal distance = x  = V₀ * √(2y/g)

Tan θ =   y / (V₀ * √(2y/g))

=> Tanθ =  √y  √g / V₀ * √2

=> √y = V₀ * √2Tanθ/√g

Squaring both sides

=> y = 2V₀²Tan²θ/g

x = V₀ * √(2y/g) = V₀ * √ ( 2 * 2V₀²Tan²θ/g²)

=> x = V₀ * 2V₀Tanθ/g

=> x = 2V₀²Tanθ/g

(2V₀²Tanθ/g , - 2V₀²Tan²θ/g)

Answer 2

Answer:

Explanation:

Horizontal speed = V₀ (Given)

Vertical Speed = 0 (Given)

Let the coordinate of stone hitting the surface = ( x , -y)

Using the second equation of motion -

S = ut + (1/2)at²

y =  (1/2)gt²

t² = 2y/g

t = √(2y/g)

Thus, the horizontal distance

= x  = V₀ × √(2y/g)

Tan θ =   y / (V₀ ×√(2y/g))

Tanθ =  √y √g / V₀×√2

√y = V₀ ×√2Tanθ/√g

On squaring both the sides, we will get -

y = 2V₀²Tan²θ/g

x =V₀ × √(2y/g)

x = V₀ × 2V₀Tanθ/g

x = 2V₀²Tanθ/g