Calculating the velocity and acceleration given vector velocity at 1 second

In summary, a physics student studying gravity on the surface of Mars throws a ball into the air at t=0s and it follows a parabolic trajectory, reaching a maximum height at t=2s. The ball's velocity at t=1s is given as v(t=1s)=2.00 m/s i + 3.72 m/s j, where i and j are unit vectors in the x and y directions. From this information, the student is asked to write the ball's velocity in vector form at t=2s and t=3s, calculate the magnitude of acceleration of gravity near the surface of Mars, determine the initial speed and launch angle, and find the distance from the launch point where the ball
  • #36
Yes. So what is the acceleration on Mars?
 
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  • #37
kuruman said:
Yes. So what is the acceleration on Mars?
So would it just be 3.72 m/s^2?
 
  • #38
That would be the magnitude. Congratulations! You have answered part (b). Now onto part (a). Can you write the velocity in vector form at t = 2 s and t = 3 s?
 
  • #39
kuruman said:
That would be the magnitude. Congratulations! You have answered part (b). Now onto part (a). Can you write the velocity in vector form at t = 2 s and t = 3 s?
Vx would stay the same and then I would add 3.72m/s to Vy. so 2.00m/s i + 7.44 m/s j
 
  • #40
At what time is this velocity? Remember that at t = 2 s the ball is at maximum height traveling horizontally.
 
  • #41
kuruman said:
At what time is this velocity? Remember that at t = 2 s the ball is at maximum height traveling horizontally.
I thought it would be a time t=2s
 
  • #42
mdavies23 said:
I thought it would be a time t=2s
It can't be at time t = 2s. Please read your post #12 and reconsider.
 
  • #43
kuruman said:
It can't be at time t = 2s. Please read your post #12 and reconsider.
Oh so Vy would be zero so the vector would look like 2.00 m/s i + 0 m/s j
 
  • #44
Right. That's at t = 2 s. What is the velocity at t = 3 s?
 
  • #45
kuruman said:
Right. That's at t = 2 s. What is the velocity at t = 3 s?
2.00 m/s i - 3.72 m/s j
 
  • #46
Very good! You are done with part (a). Time to move on to part (c). What do you need to know to find the initial speed and launch angle?
 
  • #47
kuruman said:
Very good! You are done with part (a). Time to move on to part (c). What do you need to know to find the initial speed and launch angle?
I need to know the velocity vector at t=0 then i could find the magnitude and direction o f it
 
  • #48
Right. Go for it. Remember that in 2 s vy drops from its initial value to zero.
 
  • #49
kuruman said:
Right. Go for it. Remember that in 2 s vy drops from its initial value to zero.
7.70 m/s at an angle of 75 degrees
 
  • #50
That looks right. What about part (d)?
 
  • #51
kuruman said:
That looks right. What about part (d)?
2*4 = 8m
 
  • #52
That's it. You're done with this one.
 
  • #53
kuruman said:
That's it. You're done with this one.
Thanks for everything
 

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