A skateboarder shoots off a ramp with velocity of 6.5 m/s directed at a 59 degree angle. The end of the ramp is 1.3m above the ground.
a)How high above the ground is the highest point that the skateboarder reaches?
b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
break the initial vertical velocity into vertical, and horizontal components
vertical Viv=6.5sin59
horizontalVih=6.5cos59
at the highest point, the vertical velocity will be zero.
vf=viv-g*t
t=viv/g
now use that time in
h(t)=hi+viv*t-4.9t^2
how far?
distance=vih*t