Question

A mountain biker encounters a jump on a race course that sends him into the air at 60° to the horizontal. If he lands at a horizontal distance of 45.0 m and 20 m below his launch point, what is his initial speed?

a) 12.5 m/s
b) 15.0 m/s
c) 18.7 m/s
d) 21.2 m/s

Answer 1

To find the initial speed of the mountain biker, we can use the principles of projectile motion and the equations of motion. By breaking down the initial velocity components into horizontal and vertical components, we can calculate the initial speed. The correct option is approximately option d.

Step-by-step explanation:

To solve this problem, we can use the principles of projectile motion. The initial speed of the mountain biker can be found by breaking down the initial velocity components into horizontal and vertical components. We can then use the equations of motion to calculate the initial speed.

The horizontal component of the initial velocity can be found using the equation:

vx = v * cos(θ)

where v is the initial speed and θ is the launch angle. Plugging in the values, we have:

vx = 10 * cos(60°) = 5 m/s

The time of flight can be calculated using:

t = 2 * (h - h0) / g

where h is the vertical displacement, h0 is the initial height, and g is the acceleration due to gravity. Plugging in the values, we have:

t = 2 * (20 - 0) / 9.8 = 4.082 s

The horizontal displacement can be calculated using:

d = vx * t

where d is the horizontal displacement. Plugging in the values, we have:

d = 5 * 4.082 = 20.41 m

Therefore, the initial speed of the mountain biker is approximately 20.0 m/s.