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ou push a skateboard so that it rolls down the road at a speed of 1.50 m/s. You run after the skateboard at a speed of 3.10 m/s and while still behind the skateboard you jump off the ground at an angle of 16.0° 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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ur speed with respect to skate board is (3.10-1.50) m/s = 1.60 m/s distance required so tht u land safely on the skateboard is \[\frac{u^2 sin 2\theta}{g}\]
\[d=\frac{(1.60)^2 \sin32}{g}\]
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