Jump to content
  • 0




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?

B) 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


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= 3m

B) Each boy jumped 3.75 m & 3m

Link to comment
Share on other sites

3 answers to this question

Recommended Posts

  • 0

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:





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. :-)

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

  • Create New...