Asked by p

You push a skateboard so that it rolls down the road at a speed of 1.40 m/s. You run after the skateboard at a speed of 3.80 m/s and while still behind the skateboard you jump off the ground at an angle of 30.0° above the horizontal hoping to land on the skateboard. How much distance do you need between you and the skateboard to jump and land on it?

Answered by Henry
Vo =3.80 m/s[30o].
Xo = 3.8*Cos30 = 3.29 m/s.
Yo = 3.8*sin30 = 1.9 m/s.

Y = Yo + g*Tr.
0 = 1.9 - 9.8Tr, Tr = =.194 s. = Rise time.

Tf = Tr = 0.194 s. = Fall time.

d = V*(Tr+Tf) = 1.4 * 0.388 = 0.543 m.



Answered by p
sorry it says that its the wrong answer
Answered by Henry
d1 = V*(Tr+Tf) = 1.4 * 0.388 = 0.543 m. = Distance covered by skateboard while skater is in air.

d2 = Xo*(Tr+Tf) = 3.29 * 0.388 = 1.28 m. = Distance jumped by skater.

d2-d1 = 1.28 - 0.543 = 0.734 m. Between skater and skateboard.

Answered by p
ok thanks got it
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