Projectile Motion trajectory Problem

AI Thread Summary
The discussion centers on a physics problem involving projectile motion on Planet Exidor, where a ball is thrown following a parabolic trajectory. The student seeks to determine the ball's velocity at specific times (t=0, 2, and 3 seconds) and the gravitational acceleration (g) on the planet, along with the launch angle. Initial velocity at t=1 second is given as (2.0i + 2.0j)m/s, and the vertical component of velocity is zero at the peak (t=2s), which aids in calculating g. The participant successfully deduces the gravity value and uses it to find the velocities at other times, while questioning the correctness of this approach. The thread concludes with a note on verifying the results through the instructor's resources.
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8. A Physics student on Planet Exidor throws a ball, and it follows the parabolic trajectory shown in Figure 6.8 (whille it will not be recreated on this forum, the figure shows a part of a parabolic curve which starts at the origin and peaks at t=2. It also shows the velocity vector at t=1, which will be entered later in the pst). The ball's position is shown at 1 s intervals until t=3s. At t=1 s, the ball's velocity is (2.0i + 2.0j)m/s, where i and j are unit vectors.

a. determine the ball's velocity at t=0,2,and 3 s.
b. what is the value of g on planet Exidor?
c. what was the ball's launch angle.

I know how to do parts b and c, but can't do either without a. I'm not sure what formula to use, but I do know the following:

at t=1:
v1=(2.0i + 2.0j)m/s
v1x=2.0im/s
v1y=2.0jm/s

ax=0
ay=-g m/s^2

v1x=v0x=v2x=v3x2.0im/s
x0=0m
y0=0m

Thank you in advance.
 
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Treat the vertical and horizontal components separately. Hints: What's the vertical component of velocity at the highest point? (t = 2) What about the horizontal component?
 
Figured it out. since the velocity in the y direction is 0m/s at t=2s, I was able to calculate the gravity. I then used this to find the velocities at the other times. I don't know if using part b to solve part a is correct, but I can check my instructor's webpage to see if it gets the correct answer.
 
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