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?
At what horizontal distance from the end of the ramp does he land?
1 answer:
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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