# Projectile Motion: Find Velocities and Angles

• joel amos
In summary, the conversation discusses finding the velocity of a ball thrown at an angle of 35 degrees above the horizontal, and determining its velocity after 1 second. It also addresses finding the time at which the ball will be headed at angles of 20 degrees and -20 degrees above the horizontal. There is a question regarding a discrepancy in the calculated angle at 1 second, and the conversation suggests using SUVAT equations to correctly calculate the velocity and angle.

## 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?

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
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You have the correct values for vx and vy.
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 vy(1). Then find the resultant of vx and vy(1).

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

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.

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.