Astunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp. the vertical height of the ramp is 1.5m above the cars and the horizontal distance he must clear is 22m. if the ramp is tilted upward so that the takeoff angle is 9 degrees what is the minimum speed? i got 35m but was marked wrong.

v=26.41 m/s

Step-by-step explanation:

From the Newtons laws of motions, we sew that x= (2v₁²sin∅os∅)/g where x is the horizontal distance v is the initial speed and ∅ is the launch angle.

From trigonometry we see that 2 sin∅cos∅=sin 2∅

Therefore,  x=(v²sin2∅)/g

x=22m

∅=9°

g=9.8m/s²

22m=v²×sin(2×9)/(9.8m/s²)

v²=(22×9.8)/(sin 18)

v²=697.696

v=√697.696

v=26.41 m/s

the answer is c hope it

step-by-step explanation:

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