Question

A block leaves a frictionless inclined surface horizontally after dropping off by a height h. Find the horizontal distance d where it will land on the floor in terms of h, h, and g

Answer 1

The horizontal distance, where it will land on the floor, be 2√(hH).

What is speed?

Speed is distance travelled by the object per unit time. Due to having no direction and only having magnitude, speed is a scalar quantity With SI unit meter/second.

Given parameter:

height of the inclined path = h.

Let, the mass of the block be = m.

speed of the block at the end of inclined path is v.

Then from conservation of energy, we can write:

Potential energy at the top of the inclined path  = kinetic energy at the bottom of the inclined path

⇒ mgh = 1/2 mv²

⇒ v = √(2gh)

Now, the block will take projectile motion to reach the floor and the height is H.

Time taken by the block to fall the floor: t= √(2H/g).

So, the horizontal distance d = vt = √(2gh) ×√(2H/g). = 2√(hH).

Hence, the horizontal distance, where it will land on the floor, be 2√(hH).

Learn more about speed here:

https://brainly.com/question/28224010

#SPJ1