Question 909476
The athlete left the ground with an initial velocity of magnitude {{{v}}} (in m/s), making an angle of {{{28^o}}} with the ground. The magnitude and direction of the athlete's velocity changed over time under the influence of the acceleration of gravity, {{{g=9.6}}}{{{m/s^2}}} 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*cos(28^o)}}} ,
and a vertical component of magnitude {{{v*sin(28^o)}}} .
 
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*cos(28^o)}}} 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*cos(28^o)*t=d}}} ,
and in this particular case {{{v*cos(28^o)*t=7.8}}} . 
 
FOR THE VERTICAL COMPONENT OF THE MOTION:
The vertical component of the velocity changes from {{{v*sin(28^o)}}} to zero over {{{t/2}}} seconds, at a constant rate of {{{g=9.8}}}{{{m/s^2}}} .
So, {{{v*sin(28^o)/(t/2)=9.8}}}--->{{{v*sin(28^o)=9.8*(t/2)}}}--->{{{v*sin(28^o)=4.9t)}}}--->{{{t=v*sin(28^o)/4.9)}}}
 
PUTTING IT ALL TOGETHER:
How are {{{v}}} and {{{d}}} related?
Putting together {{{v*cos(28^o)*t=d}}} and {{{t=v*sin(28^o)/4.9)}}} we get
{{{v*cos(28^o)*(v*sin(28^o)/4.9)=d}}}--->{{{d=(cos(28^o)*sin(28^o)/4.9)*v^2}}}
Substituting the approximate values {{{sin(28^o)=0.46947}}} and {{{cos(28^o)=0.88295}}} ,
to get the approximate formula {{{d=0.08460v^2}}}
 
WHAT DOES THAT TELL YOU?:
a) At this point you can use {{{d=0.08460v^2}}} to calculate {{{v}}} :
{{{7.8=0.08460v^2}}}--->{{{v^2=7.8/0.08460}}}--->{{{v^2=92.20}}}--->{{{v=sqrt(92.20)}}}-->{{{v=9.6}}} (rounded).
So, the initial velocity was {{{highlight(9.6)}}} m/s.
 
b) If an athlete jumps at the same angle, with {{{2}}} times the initial velocity, he/she will get {{{2^2=4}}} times as far.
If the athlete jumps with {{{1.05}}} times (5% more) initial velocity,
he/she will get {{{1.05^2=1.1025}}} times farther.
That would be {{{1.1025*7.8=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(0.8)}}} m longer,
or 1.10 times longer (10% longer).