- #1
DLPhysics
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Homework Statement
A rock is projected from the edge of the top of
a building with an initial velocity of 18.4 m/s
at an angle of 35◦ above the horizontal. The
rock strikes the ground a horizontal distance
of 96 m from the base of the building.
The acceleration of gravity is 9.8 m/s2 .
Assume: The ground is level and that the
side of the building is vertical. Neglect air
friction.
The problem now asks for the height of the building, the vertical component of the rock's velocity when it strikes the ground, and the magnitude of the rock's velocity when it strikes the ground.
time = 6.369258724 sec
velocity in the x = 15.07239761 m/sec
acceleration = 9.8 m/sec*2
initial velocity in the y = 0 m/sec (don't know if that's right)
Homework Equations
height = (initial velocity in the y)(time) + 1/2(acceleration)(time)*2
time = sqrt(2 * height/acc.)
final velocity in the y*2 = initial velocity in the y*2 + 2(acceleration)(height)
final velocity in the y = initial velocity in y + (acceleration)(time)
The Attempt at a Solution
I attempted to use trigonometry to find the vertical component of the rock's velocity, but that did not work, but it did work for the horizontal component, and when finding the final velocity in the y, I used the third equation listed, but that did not work either. And, I don't quite understand what the question means by "the magnitude of the rock's velocity when it strikes the ground". I'm not sure whether the initial velocity in the y is 0, 18.4, or some other number, which is probably why I'm having trouble getting the height of the building.
Can anyone help?
Damion