Physics problems need help?

Question

A stone is catapulted rightward with an initial velocity of 28.0 m/s at an angle of 46.0° above level ground. Find its horizontal and vertical displacements at the following times after launch:
(a) 1.00 s
m (horizontal)
m (vertical)
(b) 1.70 s
m (horizontal)
m (vertical)
(c) 4.1 s
m (horizontal)
m (vertical)

Answer 1

The equation necessary to find the horizontal and vertical positions of the stone are
m horizontal = (Initial horizontal velocity)(time) + (.5)(horizontal acceleration)(time)^2
m vertical = (Initial vertical velocity)(time) + (.5)(vertical acceleration)(time)^2
Since your horizontal acceleration is zero, that part goes away.
Your vertical acceleration is caused by gravity, so it should be about (-9.8m/s^2).
Your Initial horizontal velocity is (28.0 m/s)(cos (46.0)) = 19.45 m/s
Your Initial vertical velocity is (28.0 m/s)(sin (46)) = 20.14 m/s
Thus, your two equations are
m (horizontal) = (19.45 m/s) (time) and
m (vertical) = (20.14 m/s)(time) + (.5)(-9.8m/s^2)(time)^2
Plugging in your numbers, you get
a) m(horizontal) = 19.45 m; m(vertical) = 15.24 m
b) m(horizontal) = 33.1 m; m(vertical) = 25.9 m
c) m(horizontal) = 79.75 m; m(vertical) = 0 m (back on the ground)