Projectile Motion: Find Velocities and Angles

[joel amos]

Homework Statement

A ball is thrown at 12 m/s at an angle of 35 degrees above the horizontal.
(a) Find its velocity 1.0s later.
(b) At what time after it was thrown will the ball be headed at an angle of 20 degrees above the horizontal?
(c) At 20 degrees below the horizontal?

2. The attempt at a solution
For part a, I found my initial x Velocity to be about 9.8m/s and my y velocity to be about 6.88m/s. To find the velocity after a second, I simply subtracted 9.8m/s (gravity) from the initial y velocity to get -2.92m/s. Thus at one second, the y velocity was -2.92 and the x velocity was still 9.8m/s. However, I found theta to be 73.4 degrees, which much larger than the original angle. Theta at 1s should be smaller than theta at 0s. What did I do wrong?

 

[azizlwl]

For part a, I found my initial x Velocity to be about 9.8m/s and my y velocity to be about 6.88m/s. To find the velocity after a second, I simply subtracted 9.8m/s (gravity) from the initial y velocity to get -2.92m/s. Thus at one second, the y velocity was -2.92 and the x velocity was still 9.8m/s. However, I found theta to be 73.4 degrees, which much larger than the original angle. Theta at 1s should be smaller than theta at 0s. What did I do wrong
-------

You have the correct values for v_x and v_y.
Once it is airborn, neglecting air resistance, the object is subjected to gravity that will change its velocity. Clearly the gravity will not going to change the horizontal velocity which orthogonal to it.

Use SUVAT equation to find v_y(1). Then find the resultant of v_x and v_y(1).

 

[joel amos]

I thought I took care of gravity's affect on V_y after 1 second by subtracting 9.8m/s from the V_oy.

 

[Simon Bridge]

Your final y velocity is negative - so it points down.
Thus the final angle should also be negative ... arctan(vy/vx)
I suspect you got vy and vx the wrong way around.

 

[azizlwl]

The angle is changing from 20° to zero at the top.
Then slowly increasing in negative value.
The position of the ball at 1 sec. is on downward segment.