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1 month ago

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A BMX bicycle rider takes off from a ramp at a point 2.4 m above the ground. The ramp is angled at 40 degrees from the horizonta

l, and the rider's speed is 5.9 m/s when he leaves the ramp.
At what horizontal distance from the end of the ramp does he land?

1 answer:

Ostrovityanka [3.2K]1 month ago

8 0

Answer:

The BMX rider lands 5.4 meters horizontally away from the ramp's end.

Explanation:

The BMX position vector is represented as:

r = (x0 + v0 × t × cos α, y0 + v0 × t × sin α + ½ × g × t²)

Where:

r = position at time t

x0 = initial horizontal position

v0 = initial speed

α = angle of jump

y0 = initial vertical height

g = gravitational acceleration (-9.8 m/s², upward positive)

Refer to the diagram for clarity. At the landing time, the vertical coordinate of the position vector is -2.4 m, measured from the ramp's edge as the origin. Using the vertical component equation for y, one can solve for t, then substitute t to find the horizontal distance.

The vertical position equation:

-2.4 m = 0 + 5.9 m/s × t × sin 40° - ½ × 9.8 m/s² × t²

Rearranged:

0 = -4.9 t² + 5.9 t × sin 40° + 2.4

Solving this quadratic yields:

t = 1.2 seconds

Then, calculate horizontal distance:

x = 0 + 5.9 m/s × 1.2 s × cos 40° = 5.4 m

This means the BMX lands 5.4 meters from the ramp's edge.

Have a great day!

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1 month ago