Question

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 horizontal, 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?

Answer 1

Answer:

The BMX lands 5.4 m from the end of the ramp.

Explanation:

Hi there!

The position of the BMX is given by the position vector "r":

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t

x0 = initial horizontal position

v0 = initial velocity

α = jumping angle

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

Please, see the attached graphic for a better understanding of the situation. At final time, when the bicycle reaches the ground, the vector position will be "r final" (see figure). The y-component of the vector "r final" is - 2.4 m (placing the origin of the frame of reference at the jumping point). With that information, we can use the equation of the y-component of the vector "r" (see above) to calculate the time of flight. With that time, we can then obtain the x-component (rx in the figure) of the vector "r final". Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

-2.4 m = 0 m + 5.9 m/s · t · sin 40° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 5.9 m/s · t · sin 40° + 2.4 m

Solving the quadratic equation:

t = 1.2 s

Now, we can calculate the x-component of the vector "r final" that is the horizontal distance traveled by the bicycle:

x = x0 + v0 · t · cos α

x = 0 m + 5.9 m/s · 1.2 s · cos 40°

x = 5.4 m

The BMX lands 5.4 m from the end of the ramp.

Have a nice day!

Answer 2

Horizontal distance of BMX bicycle rider, who takes off from a ramp at 2.4 m above the ground land at 5.4 meters at the end .

What is projectile motion?

Projectile motion is the motion of the body, when it is thrown in the air taking the action of gravity on it.

The second equation of the motion for distance can be given as,

$s=ut+\dfrac{1}{2}gt^2$

Here, (u) is the initial body, (a) is the acceleration of the body and (t) is the time taken by it.

Given infroamtion-

The height of the ramp is 2.4 meter above the ground.

The ramp is angled at 40 degrees from the horizontal.

The speed of the rider is 5.9 m/s.

The vertical component of the velocity is,

$v_y=v_0\sin (40)$

As the initial velocity of the body is zero. Thus by the second equation of motion ,

$-2.4=5.9\sin(40)\times t-\dfrac{1}{2}\times9.8\timest^2\\t=1.2\rm s$

The horizontal distance is the product of horizontal velocity component and the time taken by it. Thus the horizontal distance is,

$x=5.9\times \cos (40)\times 1.2\\x=5.4 \rm m$

Hence, horizontal distance from the end of the ramp does the rider land is 5.4 meters.

Learn more about the projectile motion here;

https://brainly.com/question/24216590