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

AI Thread Summary
A projectile is launched at a 45-degree angle from a height of 1.2 meters with a horizontal displacement of 25.58 meters. The problem involves calculating the time of flight without considering air resistance. The initial vertical component of velocity can be derived from the launch height, and the horizontal and vertical motion equations are relevant. The calculated time of flight is approximately 2.34 seconds, pending verification of the method used. This calculation is essential for understanding projectile motion dynamics.
misterwiddle
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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.
 
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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.
 
Oh the maximum height isn't 1.2m, that's the distance above the ground it was launched at.
 
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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