Question 50

[zeina_ghou6141]

I'm a bit confused with the explanation on how horizontal range was calculated:
Horizontal range = vhorizontal x t = 4/5 x 500 x 60 = 24,000 m

Why is 4/5 in the calculation?

[admin]

The question provides information for you to construct a triangle with 4 as the horizontal, 3 as the vertical meaning (using the Pythagorean theorem) that 5 would be the hypotenuse. Cosine theta (for the horizontal component) would therefore be given by adj/hyp or 4/5.

[scmcdonald8839]

where did the 3/5 come from for the vertical velocity? I see the 3 4 5 triangle, but how did you determine the vertical velocity? I think I am having problems with my sin cos...

[kzteam5226]

the equation used in this question : t = v(vertical)/g ... I have never used. Do we use this equation simply because the units match up or is it a modification of one of the big five equations for kinematics?

[mcat_premed3832]

As you mentioned, you can use the equation because "the units match up" or, formally, dimensional analysis.

But indeed, it is one of the equations of kinematics:

Vf = Vo + at

so if Vf is 0 (for an instant at the max. height when changing directions) and we are examining the vertical acceleration -g,

0 = Vo - gt

or t = Vo/g

[admin]

Bump.

[boshenliu34457]

but the question also mentioned "in the absense of gravity", so doesn't that mean g=0?

[dltkdgn896512]

ez pz lemon squeezy