# Basic Physics Question please help

you said i left out the vertical displacement at time, so that would be y in the case of this equation y = y0 + v0y*t - 1/2*g*t^2 , and y is 53 m ?

alright so solving that equation would give me v0. Now what do i do with that? would i also need to solve for v0 in equation x=x0 + v0t?

:O I just figured something out... Don't know if im on the right track but let me know:

If i solve for v0 in equation x=x0 + v0t, then i can sub that into the other equation?

Doc Al
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Solving that equation gives you the vertical component of the initial velocity; you already solved for the horizontal component.

Alright, so i got 36.3 m/s for v0

So the horizontal component of my initial velocity is 20 m/s, and the vertical component of initial velocity is 36.6 m/s

Now i have to calculate how far is its displacement horizontally from launch pt, at the instant it achieves max height.

So i know that its v will be 0 m/s at max heigh.

Doc Al
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kara said:
So i know that its v will be 0 m/s at max heigh.
Right... the vertical component of the velocity will be zero.

and b/c im looking for how far its been displaced horizontally from the launch pt. i am looking for x?

so i can solve for t in the y=y0 + v0-1/2gt^2 equation and sub t into x=x0 +v0t equation and solve for x

Doc Al
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kara said:
and b/c im looking for how far its been displaced horizontally from the launch pt. i am looking for x?
That's right. You are looking for the value of x when y is maximum. Hint: When does it reach the maximum height?

well the max height is 53 m, and it reaches that height when v = 0.0 m/s

i plugged in all my values but get stuck at one point with a negative square root:

53 = -1/2(9.8)t^2

if i multiply both sides by 2 to get rid of the 1/2 i get

106 = -(9.8) t^2

Doc Al
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kara said:
i plugged in all my values but get stuck at one point with a negative square root:

53 = -1/2(9.8)t^2
You left out part of that equation; it should be:
53 = v0t -1/2(9.8)t^2

Where v0 is the vertical component of initial velocity that you found earlier.