Question

An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9? above the horizontal.

A.Determine x-values at each 1 s from t = 0 s to t = 6 s.

B.Determine y-values at each 1 s from t = 0 s to t = 6 s.

C.Determine vx-values at each 1 s from t = 0 s to t = 6 s.

D.Determine vy-values at each 1 s from t = 0 s to t = 6 s.

E.Determine the speed v at each 1 s from t = 0 s to t = 6 s

Answer 1

a
There is no acceleration involved on the x axis, so you only need to find the value of Vx.

Vx = Vo cos(36.9)
Vx = 39.98 m/s may as well use 40m/s.
So you got the right answer.

b
You find the initial Vy the same way.
Vy = Vo sin(36.9)
Vy = 30.02 m/s may as well use 30m/s.

But you need to add gravity to the equation, thus the position y dependant on time is
y(t) = 30m * t - 9.8m * t^2
For example, at 2 seconds it will be :
y(2) = 30 * 2 - 9.8 * 2^2
y(2) = 60 - 9.8 * 4
y(2) = 60 - 39.2
y(2) = 20.8m

c
Right, Vx remains always the same.

d
Vy slowly reduce before getting negative, as the object will fall towards the ground at some point.

Vy (t) = Vy_0 - Ay
Vy (t) = 30 - 9.8 *t

It will be reduced by 9.8m/s every second because of gravity.
For exemple, at 2 seconds:
Vy (2) = 30 - 9,8*2
Vy (2) = 30 - 19.6
Vy (2) = 10.4 m/s

e
You already have the values of Vy and Vx for each seconds between t=0 and t=6. (Vx always 40m/s).
You only need to see them as two vectors and add them together.

For exemple, at t=2 seconds,
Vx = 40m/s
Vy = 10.4 m/s
By using Pythagorean theorem:
V^2 = 40^2 + 10.4^2
V^2 = 1708.16
V = 41.33 m/s