Question

Given a total displacement of 81.1m in the positive x direction, and an inital launch angle of 45 degrees from the ground, neglecting air resistance, and without blatantly giving me the answer, what is the total time the object was in flight?

Answer 1

The range of a projectile is given by:
\[s_{x} = \frac{ 2u^{2} \sin \theta \cos \theta }{ g }\]

s_x = displacement, u = initial velocity, the angle theta is the 45 degrees and g is the acceleration due to gravity

and the total time in the air is given by:
\[t = \frac{ 2u \sin \theta }{ g }\]

so if you can solve them both for u, and then equate them to get rid of the u's, you can then solve for s_x

Answer 2

Where did the displacement formula come from? Is it a formula presented "as is" or was it derived from other formulas?

Answer 3

that's because \[s_{x} = v \times t\] so i'm basically multiplying the second equation by v on both sides, in order to change the left hand side to s_x. hence the u^2 term

Answer 4

sorry, that's not completely clear, i'm using v and u interchangably. v = u

Answer 5

also the cos theta is because its the x-component of the original velocity,

Answer 6

When you say\[s _{x}=v \times t\] you also mean to say \[s _{x}=\frac{ 1 }{ 2 }(v _{o}+ v) t\] correct? As in, v in your equation is thge average velocity?

Answer 7

since there is no air resistance the initial velocity and average velocity are the same. The velocity never changes in the x direction. So yes v_x is the average velocity

\[s_{x} = v_x \times t = \frac{ 2 v_x^{2}\sin \theta \cos \theta }{ g }\]

Answer 8

Alright. Now I'm not entirely sure where the trig functions come in. Do they come from \[v _{x}=v \times \cos \theta \]
as seen

Answer 9

yeah exactly that. The other trig function the sin one comes from the y component. This is what determines the length of time the object spent in the air. So essentially we have length of time in air \[\frac{2u \sin \theta}{g}\] multiplied by speed were traveling at along the x axis \[u \cos \theta\] equals distance. Time x speed = distance

Answer 10

Alright, solving for velocity in both of the equations I get


Correct?