A question and no idea what to do

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To determine the initial velocity of a horizontally launched projectile that falls 1.5 meters while moving 16 meters horizontally, the motion can be divided into x and y components. The horizontal motion equation is x = v_0t, while the vertical motion is described by 0 = y_0 - (1/2)gt^2, with g being the acceleration due to gravity. The user initially struggled with the problem but received guidance on using these equations, which helped them find the solution. They confirmed understanding of the equations and requested further assistance with additional questions for an upcoming test. The discussion emphasizes the importance of breaking down projectile motion into manageable components.
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 explination
 
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