physics virgin

[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?

[Fullhawking]

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.