Question

A BMX bicycle rider takes off from a ramp at a point 1.8 m above the ground. The ramp is angled at 40° from the horizontal, and the rider's speed is 6.7 m/s when he leaves the ramp.

At what horizontal distance from the end of the ramp does he land?Express your answer with the appropriate units.

Answer 1

To find the horizontal distance from the end of the ramp where the rider lands, we can use projectile motion equations. By calculating the time of flight and using it to find the horizontal distance, we can determine the answer.

Step-by-step explanation:

To find the horizontal distance from the end of the ramp where the rider lands, we need to calculate the time it takes for the rider to land. Since the rider is airborne, we can treat this as projectile motion. By using the vertical motion equation, we can find the time of flight:

`Δy = vy * t + (1/2) * ay * t2`

Since the rider lands at the same height as the takeoff point, Δy = 0. Therefore,
`t = 2 * vy / ay.`

To find vy, we can use the horizontal motion equation:

`vy = v * sin(θ)`

Plugging in the given values, we can calculate vy:
`vy = 6.7 m/s * sin(40°)`

Next, we find ay using the vertical motion equation:

ay = g

Finally, we can calculate the time of flight and use it to find the horizontal distance:

`D = v * cos(θ) * t`