- #1
AstraeaSophia
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I think I'm on the right track for this problem, but just want to re-check my work.
You are on the top of a 100m high (approximately 30 story) building. You just finished building a brick grill so the tenants of the building can have barbecues. You have one brick left and do not want have to lug it back off the roof. Luckily, you see a garbage truck with its roof open (top of trash pile is 2m above the road) coming down a road that passes by the building. The road is 30m from the building at closest approach. The truck is moving down the road with constant velocity 15m/s. You need to make sure you hit the trashpile, or someone is going to get hurt. You can throw the brick comfortably at an angle of 45 degrees above the horizontal.
a) What velocity do you need to throw the brick in order to make it to the top of the trashpile when the trash truck is 30m away?
We're going over vector components of projectile motion, so we've got some new equations in the text that describe those components. When drawing the diagram for the angle of the initial throw, I'm using that to calculate the initial velocity required to make it 30m out. I used:
vy = sin[tex]\Theta[/tex](v0)
vx = cos[tex]\Theta[/tex](v0)
[tex]\Delta[/tex]y = .5g([tex]\Delta[/tex]x2/vx2)
We've also been introduced to dot products and cross products, which I must confess I don't entirely understand yet because they're so new. Not sure if they're required for this problem yet.
I used the equation for the change in y-position and plugged in 98m for the value. Plugged in 30m for the change in x-position. My value for vx = 6.7082.
Then I used the cos[tex]\Theta[/tex] equation, plugged in my found value of vx, and eventually came out to the initial velocity needing to be 9.4868 m/s.
Is this the right approach? Just want to double-check myself before I go any further. Thanks!
Homework Statement
You are on the top of a 100m high (approximately 30 story) building. You just finished building a brick grill so the tenants of the building can have barbecues. You have one brick left and do not want have to lug it back off the roof. Luckily, you see a garbage truck with its roof open (top of trash pile is 2m above the road) coming down a road that passes by the building. The road is 30m from the building at closest approach. The truck is moving down the road with constant velocity 15m/s. You need to make sure you hit the trashpile, or someone is going to get hurt. You can throw the brick comfortably at an angle of 45 degrees above the horizontal.
a) What velocity do you need to throw the brick in order to make it to the top of the trashpile when the trash truck is 30m away?
Homework Equations
We're going over vector components of projectile motion, so we've got some new equations in the text that describe those components. When drawing the diagram for the angle of the initial throw, I'm using that to calculate the initial velocity required to make it 30m out. I used:
vy = sin[tex]\Theta[/tex](v0)
vx = cos[tex]\Theta[/tex](v0)
[tex]\Delta[/tex]y = .5g([tex]\Delta[/tex]x2/vx2)
We've also been introduced to dot products and cross products, which I must confess I don't entirely understand yet because they're so new. Not sure if they're required for this problem yet.
The Attempt at a Solution
I used the equation for the change in y-position and plugged in 98m for the value. Plugged in 30m for the change in x-position. My value for vx = 6.7082.
Then I used the cos[tex]\Theta[/tex] equation, plugged in my found value of vx, and eventually came out to the initial velocity needing to be 9.4868 m/s.
Is this the right approach? Just want to double-check myself before I go any further. Thanks!
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