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# Kinematics

## Question

Okay so here's the problem... Am I approaching this correct?

"Two boys have a hang time contest on their skateboards. At a speed of 5.0 m/s, they both jump straight up and then land on their moving skateboards. The first boy goes a horizontal distance of 7.5 m before he lands, while the second boy goes 6.0 m before he lands."

A) How long was each boy in the air?

How high did each boy jump?

This is what I did..

D= vt + 1/2 at^2

5=5x _ 1/2 0 x^2

5x/5 =7.5/5

x=1.5 s

D=vt + 1/2 at^2

6=5x+ 1/2)x^2

5x/5=6/5

x=1.2 s

A) The two boys were in the air for 1.5 s and 1.2 s .

D = vt + 1/2 at^2

D= 5(0.75) + 0 (0,75)^2

D= 3.75m

D=vt+1/2at^2

D=5(0.6)+0(0.6)^2

D= 3m

Each boy jumped 3.75 m & 3m

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I think you're on the right track, but I'm a touch confused on boy 1... d for boy 1 in the first part is 7.5 m, not 5m, correct? So that first equation should be: 7.5m=5*t+0 ==> t=1.5s

For part B, you don't know how high they go, but you do know their time in the air, so for boy A:

vi=?

vf=0

d=?

a=-9.8

t=1.5s/2 = 0.75

Solve for d using kinematic equations.

Boy 2 would then be similarly solved.

The big trick here -- labeling all your given information as horizontal or vertical motion, and keeping them separate. The only thing that correlates between the two is the time. :-)

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Yes, I meant to put 7.5 for boy 1, that was a typo.

But for part B, isnt that what I did above?

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Not quite. For part B, Boy 1, for example:

${v_f} = {v_i} + at$

${v_i} = {v_f} - at$

${v_i} = 0 - ( - 9.8{\textstyle{m \over {{s^2}}}})(0.75s)$

${v_i} = 7.35{\textstyle{m \over s}}$

Then, with vi known, solve for d.

$d = {v_i}t + {\textstyle{1 \over 2}}a{t^2}$

$d = (7.35{\textstyle{m \over s}})(0.75s) + {\textstyle{1 \over 2}}( - 9.8{\textstyle{m \over {{s^2}}}}){(0.75s)^2}$

$d = 1.18m$

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