Question

The motion of a human body through space can be modeled as the motion of a particle at the body's center of mass as we will study in a later chapter. The components of the displacement of an athlete's center of mass from the beginning to the end of a certain jump are described by the equations

xf = 0 + (10.5 m/s)(cos 18.5°)t

0.400 m = 0.610 m + (10.5 m/s)(sin 18.5°)t − ½(9.80 m/s2)t2

where t is in seconds and is the time at which the athlete ends the jump.
(a) Identify the athlete's position at the takeoff point. (Let the x and y-direction be along the horizontal and vertical direction resp

Answer 1

is that xf or f(x)?

Answer 2

probably f(x)

Answer 3

you're gonna want to solve for t(0)

Answer 4

ah ok so its quadratic eq.

Answer 5

its xf

Answer 6

its exactly how its written

Answer 7

from
0.400 m = 0.610 m + (10.5 m/s)(sin 18.5°)t − ½(9.80 m/s2)t^2 we can compute t
t1=-0.0581 sec and t2=0.738sec we sub this to
xf = 0 + (10.5 m/s)(cos 18.5°)t

Answer 8

xf = 0 + (10.5 m/s)(cos 18.5°)t
xf = 0 + (10.5 m/s)(cos 18.5°)(.738)=7.35meter

Answer 9

did you get the same answer solving the quadratic eq, for t1 and t2?

Answer 10

i dont even know how to do it lol thats why i cam on here lol

Answer 11

ok
from 0.400 m = 0.610 m + (10.5 m/s)(sin 18.5°)t − ½(9.80 m/s2)t^2 we can compute t
0=0.610-0.400 + (10.5 m/s)(sin 18.5°)t − ½(9.80 m/s2)t^2
0=0.210 + 3.332t - 4.9 t^2 or
0=0.210 + 3.332t - 4.9 t^2 or
-4.9t^2+3.332t+.0210=0
this is a quadratic equation, use the formula for quadratic equation to solve for t
where a=-4.9 b=3.332 and c=0.210 sub this to the formula
x=[-b-+sqrt(b^2 -4ac)]/2a,
x=[-(3.332)-+sqrt((3.332)^2 -4(-4.9)(0.210))]/(2)(-4.9),
x1=t1=-0.0581 sec and x2=t2=0.738sec

Answer 12

ok i read the screenshot,
from the formula x=x0+(v0)t-0.5(9.80)t^2
0.400 m = 0.610 m + (10.5 m/s)(sin 18.5°)t − ½(9.80 m/s2)t2 e see that
final take off xf=0.400m, initial take off x0=0.610m, initial velocity v0=10.5sin18.5