For a particular jump, the horizontal and vertical distances covered by a skier are in t seconds and given by x(t)=25tcosθ and y(t)=25tsinθ-4.9t^2. They are measured in meters.
1) What angle did the jumper take off if he lands at the base of a 90-meter hill after 5.1 seconds? Hint: let y(5.1)=-90
2) How much horizontal distance was covered?

Question

For a particular jump, the horizontal and vertical distances covered by a skier are in t seconds and given by x(t)=25tcosθ and y(t)=25tsinθ-4.9t^2. They are measured in meters. 
1) What angle did the jumper take off if he lands at the base of a 90-meter hill after 5.1 seconds? Hint: let y(5.1)=-90
2) How much horizontal distance was covered?

***[isuru]*** · Answer

when the  jumper touches the ground he had covered a total distance of 90m of vertical height from the top to the bottom of the mountain

***[isuru]*** · Answer

and the equation to the  vertical height is
              y(t)=25tsinθ-4.9t^2
here
  y= 90 
 t= 5.1
what will u get as the value of sin θ if u plug those values in to the above equation ?

anonymous · Answer

that is true. but how far from the start to 5.1 seconds is it?