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2D Motion (Projectile)

  • Thread starter RuthlessTB
  • Start date
  • #1
22
0

Homework Statement


A ball is kicked with initial velocity 40 m/s at an angle of 37 degree above the horizontal. What is the time (in seconds) needed to reach the maximum height?


Homework Equations


Y-axis
h= Voy t - 0.5 g t^2
Vy= Voy - g t
Vy^2= Voy^2 - 2 g h

X-axis
d= Vox t
Vx= Vox


The Attempt at a Solution


I reached to a point where I couldn't figure out a way to continue..

I used this equation
h= Voy t - 0.5 g t^2
h= (40 sin(37)) t - 5 t^2

I have 2 unknowns..
I tried to use Vy=Voy - g t
to find the time, I got so far these 2 equations
h= (40 sin(37)) t - 5 t^2
0=(40 sin(37)) - 10 t

Is this right?
 

Answers and Replies

  • #2
380
7

Homework Statement



I tried to use Vy=Voy - g t
to find the time, I got so far these 2 equations
h= (40 sin(37)) t - 5 t^2
0=(40 sin(37)) - 10 t
Is this right?
Yes, you can use either one to solve for t. If you use the upper formula, remember that you get the total flight time which you have to divide by 2 since we assume the projectile is symmetric.
 
  • #3
22
0
Thank you, I used the second equation and the answer is 2.4 s.. is it right?
 
  • #4
380
7
Thank you, I used the second equation and the answer is 2.4 s.. is it right?
Looks good.
 

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