An 65 kg Olympic runner leaps over a hurdle. The acceleration of gravity is 9.81 m/s 2 . If the runner’s initial vertical speed is 2.9 m/s, how much will the runner’s center of mass be raised during the jump? Answer in units of m.
Answers
Explanation:
think of it this way
the mass is going up at 2.7 m/s and during that time is being slowed down at a rate of 9.81 m/s^2 by gravity
eventually the vertical velocity slows to zero (and then he starts down)
how long does it take for the 2.7 to become 0 with 9.81 decelleration
the equation looks like this
v=vo+at
with vo being the initial velocity, we are looking for the "t" where the "v" equal 0, so
0=2.7 m/s + (-9.8 m/s^2) t
-2.7 m/s =-9.8 m/s^2 t
-2.7/-9.8 s=t
.276 s=t
so, it takes a little over 1/4 a second for the jumper to hit his height
how high is he at that time (or at least, how much has his center of mass moved)
the equation is
d=do+vo t + 1/2 a t^2
we'll let do be zero, ground level or more correctly initial center of mass level
d= 2.7(.276) m +.5(-9.8)(.276^2) m
d= .371 m
or a little over a foot
I may have made an arithmetic error, you should check it
If I did it right he won't clear a very high hurdle unless his original center of mass was pretty high, but these hurdlers are often long-legged
Answer:
The vertical height in which the runner’s center of mass is raised during the jump is 0.43 m.
Explanation:
The vertical height reached by the runner is calculated by applying the third kinematic equation as shown below.
Step 1: Mathematically, the formula is given as;
v² = u² - 2gh
Step 2: where;
v is the final velocity of the runner at the maximum height
u is the initial velocity of the runner
g is acceleration due to gravity
h is the height reached by the runner
at maximum height, the final velocity, v = 0
Step 3: The vertical height reached by the runner is calculated as;
0 = u² - 2gh
2gh = u²
h = u² / 2g
h = ( 2.9² ) / ( 2 x 9.8 )
h = 0.43 m
Hence, The vertical height in which the runner’s center of mass is raised during the jump is 0.43 m.
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