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Physics virgin

  • Thread starter FLANKER
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FLANKER

hey, im new to this forum and physics...i was just wondering if someone could help me with a small question i have on an assignment

A boy launches an arrow at an initial velocity of 32.5 m/s at an angle of 23.0o above the horizontal.
How high above the projection point is the arrow after 1.27 s?
 
Prijectile Motion

In this problem you can separate the two dimensions of motion, the X component and the Y component. Neglecting air resistance, the displacement in the X direction is governed by
X = V * t
t is time
V is velocity in the X direction which is constant
Displacement in the Y direction is governed by
Y = 1/2 a * t^2 + V * t
a is acceleration, in this case gravity (-9.8 M/s^2 and negative because the arrow is accelerated downward.)
t is again time
V is initial velocity in the Y direction
Because you know the initial velocity and the angle at which the arrow was fired, you can calculate the initial velocity in the X and Y directions.
Vx = 32.5 M/s * cos(23.0)
Vx = 29.9 M/s
Vy = 32.5 M/s * sin(23.0)
Vy = 12.7 M/s
To answer this problem, all you really need in the Y direction.
Y = 1/2 (-9.8 M/s^2) * (1.27 s)^2 + 12.7 M/s * 1.27 s
Y = 8.23 M
So at time = 1.27 sec, the arrow is at a height of 8.23 M.
 

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