Initial velocity given time and angle

In summary: If ##v_y = 13.5 \sin(40) \ m/s##, then how long does that take to come down?Hint: you can ignore ##v_x##.
  • #1
LR5
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Homework Statement
Ball 1 is thrown straight up in the air and comes back down after 2.75 seconds. Same ball (Ball 2) is thrown at a 40 degree angle and takes the same time to come down. What is Ball 2's initial velocity?
Relevant Equations
Equations of motion
From Ball 1 I can can determine it's initial velocity and then maximum height. I'm not sure how this is relevant, as I cannot find a formula using angle, time, height and velocity. I have found h=V(squared) x sin (squared) Theta / 2g and also t=2Vsin Theta/g. These give similar, but different results, though and the second one can be used without any information from Ball 1, so why include it? I think I am mixing up my formulas, but have no idea where to go.
 
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  • #2
LR5 said:
Homework Statement:: Ball 1 is thrown straight up in the air and comes back down after 2.75 seconds. Same ball (Ball 2) is thrown at a 40 degree angle. What is Ball 2's initial velocity?
Homework Equations:: Equations of motion

From Ball 1 I can can determine it's initial velocity and then maximum height. I'm not sure how this is relevant, as I cannot find a formula using angle, time, height and velocity. I have found h=V(squared) x sin (squared) Theta / 2g and also t=2Vsin Theta/g. These give similar, but different results, though and the second one can be used without any information from Ball 1, so why include it? I think I am mixing up my formulas, but have no idea where to go.

All you've said is that ball 2 is thrown at a 40 degree angle. How can you deteremine its velocity from this alone? Its initial velocity could be anything. There must be more information. Does it, for example, also take 2.75 seconds to come down?
 
  • #3
PeroK said:
All you've said is that ball 2 is thrown at a 40 degree angle. How can you deteremine its velocity from this alone? Its initial velocity could be anything. There must be more information. Does it, for example, also take 2.75 seconds to come down?
Sorry, yes it also takes 2.75 seconds...
 
  • #4
LR5 said:
Sorry, yes it also takes 2.75 seconds...

So, what do you know about components of position, velocity and acceleration?
 
  • #5
Vsquared=V(initial)squared + 2ax?
 
  • #6
LR5 said:
Vsquared=V(initial)squared + 2ax?
Position, velocity and acceleration are vectors and have components in each direction. Moreover, the equations of motion apply in each direction. In two or three dimensional problems, therefore, you can look at the motion in each direction separately.

In this case, you can look at motion in the y-direction (vertical) for ball 2 - this should be the same as for ball 1.

Have you seen notation like ##v_x = v \cos \theta## and ##v_y = v \sin \theta##?
 
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  • #7
So, I could just take the velocity of Ball 1 as the average velocity for Ball 2? (Vy = 13.5 sin 40)?
 
  • #8
LR5 said:
So, I could just take the velocity of Ball 1 as the average velocity for Ball 2? (Vy = 13.5 sin 40)?

If ##v_y = 13.5 \sin(40) \ m/s##, then how long does that take to come down?

Hint: you can ignore ##v_x##.
 
  • #9
Lol, ok. That's only half the time. Thanks, I have been making it a lot more complicated than it is.
 
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