A mountain biker encounters a jump on a race course that sends him into the air at 35,2° to the horizontal. If he lands at a horizontal distance of 30,1 m and 14,7 m below his launch point, what is his initial speed?

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t Equation (1)

Where:

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m/s

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

vfy= v₀y -gt Equation (3)

Where:

y: vertical position in meters (m)

y₀ : initial vertical position in meters (m)

t : time in seconds (s)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

Data

α₀=35,2° angle v₀ above the horizontal

x = 30.1 m

y= 0

xi = 0

y₀ = 14,7 m

g= 9.8 m/s²

x-y components of v₀

v₀x= v₀*cos(35,2)°

v₀y= v₀*sin(35.2)°

Problem development

We replace data in the equation (1)

x = xi + v₀x*t

30.1 = 0 + v₀*cos(35,2)°(t )

t = (30.1) / (v₀*cos(35,2)°)

t = (36.84) /(v₀)

We replace data in the equation (2) ,

v₀y= v₀*sin(35.2)°, t = (36.84) /(v₀)

y= y₀+(v₀y)*t - (1/2)*(g)*t²

0 = 14,7 + ( v₀*sin(35.2)°)* (36.84)/(v₀)) - (1/2)*(9.8)*(36.84)²/(v₀)²

We eliminate v₀ in the term second of the rihgt

(1/2)*(9.8)*(36.84)²/(v₀)² = 14.7 + sin(35.2)° (36.84)

6650.209 = 35.9357 (v₀)²

(v₀)² = (6650.209) / (35.9357)

(v₀)² = 185.0585

[tex]v_{o} =\sqrt{185.0585}[/tex]

v₀ = 13.6 m/s