physics help?

Question

I need help with part b, I know part a.

An athlete executing a long jump leaves the ground at a 35 degree angle & travels 6.8m.
a) what was the takeoff speed? 8.3.
b)If this speed were increased by just 2%, how much longer would the jump be?.

So far this is what I did but I got the wrong answer....
8.3 x .02 = .166 --> V0=8.3 + .166=8.466..
Vx0=8.466cos(35)=6.93
Vy0=8.466sin(35)=4.85

y=y0 + Vy0t + .5(-9.8)(t^2)


help am i doing it right?

Answer 1

Your first part is not correct.

First, you can write these equations:

V_y=V0sin(35)
V_x=V0cos(35)

From Vy, you can find the hang time of the jump:

t=2*V_y/g

The distance of the jump is:

d=6.8=Vx*t=V0cos(35)*t=V0cos(35)*2*V0sin(35)/g
d=6.8=V0^2*sin(70)/g (I used the trig identify 2*sin(theta)*cos(theta)=sin(2*theta) here)

Put all those equations together to solve for V0:

V0=sqrt[(d*g)/sin(2*theta)]=8.4 m/s

Now for part 2, you have all the formulas already:

Let V1=new take off velocity = 1.02*V0:

Then Vx1=1.02*Vx, Vy1=1.02*Vy. Solve for the new hang time and the new distance. In this case, you end up with a jump that is 7.07m, which is 4% longer than the original jump.

Answer 2

a) range = V^2 2sin2(theta) / 9.8

6.8 = V^2 sin2(35) / 9.8

66.64= V^2

V^2 = 70.91

V = 8.4m/s

.02 ( 8.4) + 8.4 = 8.568m/s

find the component of the velcity
sin(35)8.568 = 4.91m/s

Vf = at + Vi
-4.91 = (-9.8)t + 4.91
-9.82 = (-9.8)t
t = 1s