Review During the 2014 Olympic games, a snowboarder rose 5.8 m vertically above the rim of the half-pipe.

Question

With what speed did he launch from the rim?

Answer 1

For uniform accelerated motion we have equation V_fin² -V_in²=2aD, in this case a=g=-9.8 m/s² , D=5.8 m, and V_fin=0, as the velocity at upper point of trajectory; only unknown is V_in; then 0 -V_in²=2*(-9.8 m/s²)*(5.8 m) that gives

V_in=10.66 m/s

Answer 2

we can solve this with the equation of motion in the y-direction

y = y0 +v0*t + 1/2 * g * t^2

Applying our knowns at the point of the top of the ramp we get the following

5.8[m] = v0*t + 1/2 * (9.81 [m/s^2]) *t^2

We need to find the time of flight to get the velocity. First derivative test results in the following

0 = v0 + gt --> t = v0/g

inserting this equation into what we know about the path we get the velocity to perform this height

5.8[m] = v0^2/g +v0^2/(2g) --> 2g* 5.8[m] =3( v0^2) -->

v0 = sqrt(2/3*5.8[m]*9.81 [m/s^2]) = 6.158 m/s