An athlete executing a long jump leaves the ground at a 20° angle and travels 5.80 m.?

Question

An athlete executing a long jump leaves the ground at a 20° angle and travels 5.80 m.
(a) What was the takeoff speed?
m/s
(b) If this speed were increased by just 5.0%, how much longer would the jump be?
m

Answer 1

a) I'm going to take a shortcut here and give the range formula:

d = v^2 sin(2 x angle) / g
rearranging for v
v = sqroot(d x g / sin (2 x angle))
v = sqroot(5.8 m x 9.8 m/s^s / sin (40))
v = 9.40 m/s

b) Make the speed 1.05 x 9.40 m/s = 9.87 m/s
repeat above and you get

d = v^2 sin(2 x angle) / g
d = (9.87 m/s)^2 * sin(40) / 9.8 = 6.39 m

6.39 m - 5.8 m = .59 m farther

Answer 2

Assume 5.80m is horizontal

There is a range formula you can use but teachers often don't want you to use it, so I'll give it an end. Lets say time taken is t

vx = 5.80/t

delta y = 0 so 0 = viyt + 1/2 g t^2 or t = 2viy/9.8

so vx = 5.8*9.8/(2viy) or vxviy = 5.8*9.8/2 = vcos20vsin20

solve v^2 = 5.8*9.8/(2*cos20*sin20) for v

The formula is Range = v^2sin(2theta)/g

Increasing v by 5% means range increases by 10%

So 5.80 + .580

To show it, add 5% to v found earlier and find new range

Answer 3

Long Jump Ground

Answer 4

a)difficult
got equations s=.5at^2+vit and vf^2=vi^2+2as
first one is for horizontal
5.8=(cos20vi)2t (first part of equation canceling because a=0)
simplifies to vit=3.09
now for vertical
s=.5(9.81)t^2 (second part of equation canceling because v is 0)
simplifies to s=4.905t^2
0^2=(sin20vi)^2+2(-9.81)s
simplifies to s=.005962vi^2
k. got three equations
vit=3.09
s=4.905t^2
s=.005962vi^2
set second two equal to eachother...
4.905t^2=.005962vi^2
get 822.687t^2=vi^2
get 28.683t=vi
put into first equation...
28.683t^2=3.09
t=.328sec
put back into second equation to get s=.528sec
put into third to get vi=9.41m/s

b)a little easier...
9.41(1.05)=9.88
((sin20(9.88))^2)/(2(9.81))=s
s=.582
sqrt(.582/(.5(9.81)))=t
t=.344
s=cos20(9.88)2(.344)
s=6.396
6.396-5.8=difference
difference=.596m