Solving Projectile Problem: Initial Speed 100m/s, Object 5m Away 2m High

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To solve the projectile problem with an initial speed of 100 m/s, the goal is to determine the launch angle that allows the object to just clear a 2 m high obstacle located 5 m away. The equations used involve vertical motion calculations, with the vertical position expressed as y = (Vy + (1/2)(-9.8)(t^2)) and time calculated as t = 5/(Vx). The horizontal component of velocity, Vx, is derived from the initial speed and the angle, leading to the equation Vx^2 = 100 - Vy^2. The discussion highlights the complexity of the calculations and suggests using the function Y = f(x) to simplify solving for the angle. Overall, the participants are seeking a more efficient method to find the correct launch angle.
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Im given the initial speed (100m/s) of an object launched at an unkown angle, it needs to just barely go over an object 5m away 2m high.

Ive made several attempts at this, all of which ending up in the trash
my most recent try is

y=(Vy + (1/2)(-9.8)(t^2)
t = 5/(Vx)

Then used Vx^2 = 100 - Vy^2
and subsituted it all in, and i come out with a big mess to solve, so I've assumed this is not the correct (or best) way to do this

Sorry for not going into more detail, but you should be able to get the idea
 
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BTW I am trying to find the angle...and the 100m/s is the Hypot. componet of the velocity
 
try putting Y = f(x), so you can plug your X, Y and Vo values, and solve for theta.
 
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