SOLUTION: I need someone who is good in physics i need help please .i do not understand this question. An athlete executing a long jump leaves the ground at 28.0 angle and travels 7.80 m

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Question 909476: I need someone who is good in physics i need help please .i do not understand this question.
An athlete executing a long jump leaves the ground at 28.0 angle and travels 7.80 m a)What was the take off speed? b)If this speed were increased by just 5.0%,how much longer would the jump be?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The athlete left the ground with an initial velocity of magnitude v (in m/s), making an angle of 28%5Eo with the ground. The magnitude and direction of the athlete's velocity changed over time under the influence of the acceleration of gravity, g=9.6m%2Fs%5E2 downwards, until the athlete landed on the ground , t seconds after leaving the ground.
The velocity can be decomposed into a horizontal component of magnitude v%2Acos%2828%5Eo%29 ,
and a vertical component of magnitude v%2Asin%2828%5Eo%29 .

FOR THE HORIZONTAL COMPONENT OF THE MOTION:
We assume that the athlete is jumping over a perfectly horizontal surface, and that air resistance does not really affect him/her at the speeds he/she is capable of attaining.
That makes his horizontal speed constant at v%2Acos%2828%5Eo%29 for the t seconds the athlete is airborne.
As a consequence of our assumptions, we know that
the athlete's initial horizontal velocity (in m/s)
times the time in the air (t , in seconds)
equals the horizontal displacement, d in meters (7.8 m, in this case).
So, in general v%2Acos%2828%5Eo%29%2At=d ,
and in this particular case v%2Acos%2828%5Eo%29%2At=7.8 .

FOR THE VERTICAL COMPONENT OF THE MOTION:
The vertical component of the velocity changes from v%2Asin%2828%5Eo%29 to zero over t%2F2 seconds, at a constant rate of g=9.8m%2Fs%5E2 .
So, v%2Asin%2828%5Eo%29%2F%28t%2F2%29=9.8--->v%2Asin%2828%5Eo%29=9.8%2A%28t%2F2%29--->v%2Asin%2828%5Eo%29=4.9t%29--->t=v%2Asin%2828%5Eo%29%2F4.9%29

PUTTING IT ALL TOGETHER:
How are v and d related?
Putting together v%2Acos%2828%5Eo%29%2At=d and t=v%2Asin%2828%5Eo%29%2F4.9%29 we get
v%2Acos%2828%5Eo%29%2A%28v%2Asin%2828%5Eo%29%2F4.9%29=d--->d=%28cos%2828%5Eo%29%2Asin%2828%5Eo%29%2F4.9%29%2Av%5E2
Substituting the approximate values sin%2828%5Eo%29=0.46947 and cos%2828%5Eo%29=0.88295 ,
to get the approximate formula d=0.08460v%5E2

WHAT DOES THAT TELL YOU?:
a) At this point you can use d=0.08460v%5E2 to calculate v :
7.8=0.08460v%5E2--->v%5E2=7.8%2F0.08460--->v%5E2=92.20--->v=sqrt%2892.20%29-->v=9.6 (rounded).
So, the initial velocity was highlight%289.6%29 m/s.

b) If an athlete jumps at the same angle, with 2 times the initial velocity, he/she will get 2%5E2=4 times as far.
If the athlete jumps with 1.05 times (5% more) initial velocity,
he/she will get 1.05%5E2=1.1025 times farther.
That would be 1.1025%2A7.8=8.6 (rounded).
So, jumping with 5% higher initial velocity, the athlete will travel 8.7 meters.
That means that his/her jump would be
8.6 m - 7.8 m = highlight%280.8%29 m longer,
or 1.10 times longer (10% longer).