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Projectile Motion (Find T with ΔDx, ΔDy, θ, and A?)

  • #1

Homework Statement



A projectile is launched at a 45 degree angle 1.2m off the ground. Its horizontal displacement is 25.58m. Assuming no wind resistance or friction or anything like that, how long did the launch take?

horizontal:
ΔDx=25.58m

Vertical:
ΔDy=1.2m
a=9.8m/s^2

θ, or Angle = 45 degrees

Homework Equations



I'm not sure so I'll just list a bunch I think could be relevant.

Standard BIG 4 equations.

V=D/T

a=(Vf-Vi)/t

d=([Vf+Vi]/2)t

The quadratic formula.

The Attempt at a Solution



I don't even know where to begin to be honest, I don't know any speed/velocity.
 

Answers and Replies

  • #2
165
1
You know that the angle of launch is 45 degrees, which indicates that both the horizontal and vertical components of the initial velocity are equal.

With the information that you have, you can calculate the initial vertical component of velocity if 1.2m is the maximum height.
 
  • #3
Oh the maximum height isn't 1.2m, thats the distance above the ground it was launched at.
 
  • #4
165
1
Well then, forget the second part - you can still use the first piece of information I gave you. If my method is correct, then my time is 2.34s. Can you verify that?
 

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