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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/...Asked by Anonymous
You push a skateboard so that it rolls down the road at a speed of 1.20 m/s. You run after the skateboard at a speed of 3.30 m/s and while still behind the skateboard you jump off the ground at an angle of 28.0° 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?
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Answered by
bobpursley
If you do it in relative speed, it is easy..
relative speed=3.3-1.2=2.1 m/s
horizontal distance=2.1cosTheta*time
vertical time in air:
hf=hi+2.1sin28 t-4.8t^2
0=0+t(2.1sin28-4.8t)
t=2.1sun28/4.8 Now put that time in the horiztonal distance, and solve for distance.
relative speed=3.3-1.2=2.1 m/s
horizontal distance=2.1cosTheta*time
vertical time in air:
hf=hi+2.1sin28 t-4.8t^2
0=0+t(2.1sin28-4.8t)
t=2.1sun28/4.8 Now put that time in the horiztonal distance, and solve for distance.
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