Basic Projectile Motion Problem

  • Thread starter Jeff231
  • Start date
  • #1
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Homework Statement



A physics student on PLanet Exidor throws a ball, and it follows hte parabolic trajectgory shown in the figure. The ball's position is shown at 1.00 intervals until t=3.00. At t=1.00, the ball's velocity is v=(2.00i + 2.00j)[m/s].

physicsa.jpg


a. Determine the ball's velocity at t=0, 2.00, and 3.00.
b. What is the value of g on the Planet Exidor?
c. What was the ball's launch angle?

Homework Equations



[tex]
v = v_0 + a t
[tex]

[tex]
x = x_0 + v_0 t + (1/2) a t^2
[tex]
[tex]
v^2 = v_0^2 + 2 a \Delta x
[tex]



The Attempt at a Solution



I was able to find at 1.00 that the ball's acceleration in the x direction = 2.00[m/s^2]. I'm not really sure how to find the velocity at t=0 since I think it can't be zero. I'm pretty if I can figure out the max height I can figure out what g is. But I can't find max height.

Thanks.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Kurdt
Staff Emeritus
Science Advisor
Gold Member
4,826
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With all projectile motion problems you need to consider both components of motion. There should be no acceleration in the x direction. At the top of the parabola what do you expect the y component of velocity to be? This will help you with what you know about t = 1 to set up some simultaneous equations and solve the problem.
 
  • #3
92
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There should be no acceleration in the x direction.

It will really help to rewrite

[tex]
v = v_0 + at = v_0 - gt
[/tex]

where i've defined g downwards as positive as two equations then...

[tex]
v_x = ... \quad \& \quad v_y = v_y(t)=...
[/tex]
 
  • #4
6
0
Thanks for the help. It was stupid of me to think of ax instead of just thinking g the whole time. This clarified things a bunch, thanks!
 

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