Question

A motorcycle stuntman must jump from a ramp to a platform. The platform is located 15M away and is 2M higher then the tip of the ramp. The ramp is at a 35Degree angle. At what velocity must he travel as he leaves the ramp in order to reach the platform. Express in KM/h

Answer 1

v(x)=v• cosα, v(y) =v•sinα.
s=v(x) •t => t=s/v(x)=s/ v• cosα
h=v(y)•t-gt²/2=( v•sinα•s/v•cosα) – (g•s²/2•v²•cos²α).
Solve for “v, and obtain:

v=sqrt{g•s²/2cosα(s•sinα - h•cosα)=
13.82 m/s= 50 km/h

Answer 2

Im still confused on how you solved for V, would you mind plugging in the numbers for me? im completly lost

Answer 3

h=v(y)•t-gt²/2=( v•sinα•s/v•cosα) – (g•s²/2•v²•cos²α),
2•v²• cos²α•h=2•v²•s•sin α• cosα - g•s²,
g•s²=2•v²•s•sin α• cosα - 2•v²• cos²α•h,
g•s²= 2•v²• cosα(s•sin α-h•cosα).

v=sqrt{g•s²/2cosα(s•sinα - h•cosα)=
sqrt{9.8•15²/2•cos35(15•sin35-2•cos35)}=
13.82 m/s= 50 km/h

Answer 4

thanks!