A block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane. (a) What is its velocity when it reaches the top of the plane? (b) How far horizontally does it land after it leaves the plane?

A.) 8 m/s

B.) 7.0 m

Explanation:

Given that a block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane.

(a) What is its velocity when it reaches the top of the plane?

Since the plane is frictionless, the final velocity V will be the same as 8 m/s

The velocity will be 8 m/s as it reaches the top of the plane.

(b) How far horizontally does it land after it leaves the plane?

For frictionless plane,

a = gsinø

Acceleration a = 9.8sin28

Acceleration a = 4.6 m/s^2

Using the third equation of motion

V^2 = U^2 - 2as

Substitute the a and the U into the equation. Where V = 0

0 = 8^2 - 2 × 4.6 × S

9.2S = 64

S = 64/9.2

S = 6.956 m

S = 7.0 m

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