Projectile Motion - hammer sliding off a roof

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SUMMARY

The discussion centers on calculating the horizontal distance a hammer travels after sliding off a roof inclined at 40 degrees, with an initial velocity of 7 m/s and a height of 14 m. Participants emphasize the importance of decomposing the initial velocity into its x and y components using trigonometric functions. The correct approach involves applying kinematic equations to determine the time of flight and subsequently the horizontal distance traveled. The key equations utilized include the horizontal motion equation \(x = V_0 \cos(\theta) t\) and the vertical motion equation \(y = V_{0y} t - \frac{1}{2} g t^2\).

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Nickg140143
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Homework Statement


A hammer slides off the roof of a house that slopes downward at 40 degrees. The edge of the roof is 14m avove the ground and the hammer leaves the roof at 7m/s.

How far from the edge of the roof does the hammer strike the ground?

Homework Equations


x direction
Vx=V0x

x=V0cosθy direction
Vy=V0y−gt

y=V0yt−12gt2

Vy^2=V0y^2−2gy

The Attempt at a Solution


just looking for a quick bit of help here. You can see some of the work I've done in the attached image.

My initial thought was to first decompose the initial velocity vector into its components so I could go ahead and use those in my calculations, which I did, and they seem to check out when compared with the original problems statement on velocity.

Since I'm trying to find the horizontal distance from the roof ledge, I assumed that using this would be the best bet.
(x-x_0)=V_0\cos{\theta}(t)

noticing that I need time for this, I decided to use the formula to solve for t
y={V_{0y}}^2t+\frac{at^2}{2}
Where I would let y be -14, since my displacement will be from 14m above 0m (0-14)

however, as you'll see from my work found in the attachment, it looks like I'll keep ending up with complex numbers for time.

I'm wondering whether my assumed initial variables are wrong, or if I've once again made a mathematical blunder?

Any help would be great.
 

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Well i miss read the problem at the start and had a few paragraphs describing what's going on, but oh well

The roof is on an incline of 40 degrees and your magnitude of velocity is 7m/s

you need to break this up into it's x and y components using trig

x = vsin(theta)

y = vcos(theta)

Since there are no external forces in the x direction the velocity is constant, but in the y direction there is a gravity.

So by using your kinematic equations and knowing the height of the roof and what gravity is you need to solve for the time it takes for the hammer to reach the ground with your initial y velocity and final y velocity which is zero.

You know g, d, vy1 and vy2 and your x velocity

I missed your picture at first and it looks good to me.

good luck
 

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