# A block starts at rest and slides down a frictionless track. At what initial height h above the ground is the block released? The acceleration of gravity is 9.81 m/s2. It leaves the track...

A block starts at rest and slides down a frictionless track. At what initial height h above the ground is the block released?

The acceleration of gravity is 9.81 m/s2. It leaves the track horizontally, striking the ground 3.8 meters away. The mass of the block is 467g. The block leaves the track at a height of 2.4 m.

At what speed does it leave the track? At what speed does it hit the ground?

### 1 Answer | Add Yours

The key to solving this problem is to determine how long the block was in the air before it hits the ground. It is assumed in problems of this type that motion is the x- and y-directions are independent and that air resistance is ignored.

As the block leaves the table it has zero velocity in the y-direction. Due to the acceleration of gravity it will start accelerating is it falls.

The time it takes to fall is related to the height and the initial velocity by the kinematic equation:

delta y = Vit + 1/2 gt^2 where delta y is the height, Vi is the initial velocity, g is 9.81 m/s/s, and t is the time.

In your problem:

2.4 m = 0 + 1/2 * 9.81 * t^2 and solve for t.

t = 0.7 seconds.

Now you know that the block landed 3.8 m from the edge of the table while traveling horizontally for 0.7 s.

d = vt, so v = d/t = 3.8m/0.7s = 5.43 m/s as it leaves the edge of the table in the x-direction.

The velocity in the y-direction is given by:

Vf^2 = Vi^2 + 2g * delta-y

Vf^2 = 0 + 2 * 9.81 * 2.4

Vf = 6.86 m/s in the y-direction.

The resultant velocity is: (6.86^2 + 5.43^2)^.5 = 8.75 m/s

The angle at which the block hits the ground can be found using inverse tangent

tan-1 (6.86/5.43) = 51.6 degrees down from horizontal.

**Sources:**