An archer standing on a 15 degree slope shoots an arrow at an angle of 26 degrees above the horizontal. How far below its original point of release does the arrow hit if it is shot with a speed of 23 m/s from a height of 1.98 m above the ground?

12000j is work done

no. dot product   =0 when vecs at right angles

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer