# Answer to Question #1371 in Mechanics | Relativity for frank

Question #1371

I need the formula for calculating speed in a parabolic trajcetory with no air resistance. I need to reach a range of 20meters from a 45 degree angle. what speed do I need to be travelling at? I need the formula for future reference. Thank you.

Expert's answer

Denote the angular of the speed vector to the horizontas α, the horizontal and vertical projections of theinitial speed as Vox and V

V

The equations of the motion:

V

H =H

S = V

For the case of α = 45, S = 20:

As sin(45) = cos(45) = √2/2 , Vox= √2/2 Vo = Voy

S = √2/2 Vo t = 20

t = 20√2 /Vo

As moving upward lasts only a half of motion time and at the highest point Vy =0,

V

√2/2 Vo =10* 20√2 /2Vo

Thus,

Vo

Vo = 10√2

_{oy}respectively. V_{ox}= V_{o}cos(α)V

_{oy}= V_{o}sin (α)The equations of the motion:

V

_{y}= V_{oy}- gtH =H

_{o}+ V_{oy}t - gt^{2}/2S = V

_{ox}tFor the case of α = 45, S = 20:

As sin(45) = cos(45) = √2/2 , Vox= √2/2 Vo = Voy

S = √2/2 Vo t = 20

t = 20√2 /Vo

As moving upward lasts only a half of motion time and at the highest point Vy =0,

V

_{oy}= gt/2√2/2 Vo =10* 20√2 /2Vo

Thus,

Vo

^{2}= 10* 20 = 200Vo = 10√2

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