Question

A student releases a block of mass m at the top of a slide of height h1. the block moves down the slide and off the end of the table of height h2, landing on the floor a horizontal distance d from the edge of the table. Friction and air resistance are negligible. The overall height H of the setup is determined by the height of the room. Therefore, if h1 is increased, h2 must decrease by the same amount so that the sum h1+h2 remains equal to H. The student wants to adjust h1 and h2 to make d as large as possible.
A) 1) Without using equations, explain why making h1 very small would cause d to be small, even though

h2 would be very large?
2) Without using equations, explain why making h2 very small would cause d to be small, even though
h1 would be large

B) Derive an equation for d in terms of h1, h2, m, and physical constants as appropriate.

Answer 1

Answer:

Explanation:

(a) The launch velocity would be very small as a result the height would be very small.

(b) Because the time of flight would be very small and the object wouldn't have much chance to travel horizontally before impacting vertically.

(c) mgH= 1/2 mv^2+ mg (h2)

mg(h1)+ mg(h2)= 1/2 mv^2+ mgh2

mg(h1) = 1/2 mv^2

root 2g(h1)= v(b)

In the second case, the object will travel h2= 1/2 gt^2

                                                                    2h2= gt^2

                                                                     2h2/g = t^2

                                                                     root 2h2/g = t

d= v*t = 2 root h1h2    

Answer 2

Answer:

B)   d = √  ( 4 h₂ ( H - 2h₂))

Explanation:

A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance

2) If the height is very small downwards, it touches the earth a little and the horizon is small,

B) to find an equation for horizontal distance (d)

We must maximize the speed at the bottom of the slide let's use energy

Starting point Higher

          Em₀ = U = m g h₁

Final point. Lower (slide bottom)

            Emf = K + U = ½ m v² + m gh₂

As there is no friction the energy is conserved

             mgh₁ = ½ m v² + mgh₂

             v² = 2 g (h₁-h₂)

This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)

The distance traveled when leaving the table can be searched with kinematics, projectile launch

           x = v₀ₓ t

          y = $v_{oy}$ t - ½ g t²

The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero

         t = √ 2h₂ / g

We substitute in the other equation

         d = √ (2g (h₁-h₂))  √ 2h₂ / g

         d = √ (4 h₂ (h₁-h₂))

         H = h₁ + h₂

         h₁ = H -h₂

         d = √  ( 4 h₂ ( H - 2h₂))