A question and no idea what to do

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The discussion focuses on calculating the initial velocity of a horizontally launched projectile that falls 1.5 meters while traveling 16 meters horizontally. The key equations used are x = v_0t for horizontal motion and 0 = y_0 - (1/2)gt^2 for vertical motion, where g is the acceleration due to gravity (9.8 m/s²). The user successfully applied these equations after clarification on separating the motion into x and y components, confirming that the initial vertical velocity (v_{y0}) is zero for horizontal launches.

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ouse
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determine the inital velocity of a projectile that is luncked horizantaly and falls 1.5m while moving 16m horizantaly
i tried everything all i figerd out is that they r displacmednts the accelaration is 9.8 and nothing i don't have enogh info to do anything
please help me
 
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Divide the motion into x and y components and deal with them as if you were dealing with one-dimensional motion.
 
can u please elaborate
 
and any way that doesn't work becuse i have no angles to detrime wat the x and y components are
 
that would only work if these were vectors not displacment
 
Well...
x = v_0t
0 = y_0 -\frac{1}{2}gt^2
Does that help at all?
 
yeah it helped i acctualy figerd the question thanks sooo much
 
now can u please look at my other thread and please ohh please help with question 1 b,c and q 2
 
ouse said:
yeah it helped i acctualy figerd the question thanks sooo much
Glad to hear that :).

Did you understand where the equations I gave came from?
 
  • #10
ummm well yeah for one of them
 
  • #11
u assumed viy is 0 and took it our of the equation right
 
  • #12
The full equations:
x = x_0 + v_{x0}t + \frac{1}{2}a_xt^2
y = y_0 + v_{y0}t + \frac{1}{2}a_yt^2
Let x_0 = 0. As the projectile's fired horizontally: v_{y0} = 0 thus v_{x0} = v_0. No force affects the projectile in x-direction, so: a_x = 0. In y-direction: a_y = -g (- as it's downward). The final altitude of the projectile is y = 0.

Hence
x = v_0t
0 = y_0 -\frac{1}{2}gt^2
 
  • #13
i was woundring if u could help me with the other thread please help me i really need it i have a test i need to pass
 
  • #14
thx for the explanation
 

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