Question

You are fixing the roof of your house when a hammer breaks loose and slides down. the roof makes an angle of 45 ∘ with the horizontal, and the hammer is moving at 5.5 m/s when it reaches the edge. assume that the hammer is moving from the top of the roof to its right edge. part a what is the horizontal component of the hammer's velocity just as it leaves the roof?

Answer 1

The horizontal component of the hammer's velocity just as it leaves the roof is 3.89 m/s.

Step-by-step explanation:

To find the horizontal component of the hammer's velocity just as it leaves the roof, we can use trigonometry. The horizontal component of the velocity is equal to the initial velocity multiplied by the cosine of the angle the roof makes with the horizontal.

Given that the hammer is moving at 5.5 m/s and the angle of the roof is 45 degrees, we can calculate the horizontal component of the velocity as follows:

Horizontal velocity = 5.5 m/s * cos(45 degrees) = 5.5 m/s * 0.7071 = 3.89 m/s

Therefore, the horizontal component of the hammer's velocity just as it leaves the roof is 3.89 m/s.

Answer 2

The horizontal component of the hammer's velocity is 3.9 m/s.

The hammer falls with a velocity v along the roof which is at an angle θ with horizontal.

Resolve the velocity v into two components- horizontal component vcosθ and the vertical component v sinθ.

Calculate the horizontal component of the hammer's velocity, by substituting 5.5 m/s for v and 45° for θ.

`vcos\theta = (5.5 m/s)(cos45^o)=3.889 m/s`

The horizontal component of the hammer's velocity is 3.9 m/s