A projectile, fired with unknown initial velocity, lands 18.4 s later on the side of a hill, 2740 m away horizontally and 469 m vertically above its starting point. (Ignore any effects due to air resistance.)

Question

(a) What is the vertical component of its initial velocity?
m/s

(b) What is the horizontal component of its initial velocity?


m/s

(c) What was its maximum height above its launch point?
m

(d) As it hit the hill, what speed did it have and what angle did its velocity make with the vertical?
v = m/s
θ = °

Answer 1

vertical:
469=Viv*t -1/2 g t^2 solve for viv.

horizontal
2740=vih*t solve for vih

max height: mgh=1/2 m viv^2
solve for h.
vat wall= sqrt(vih^2+vivf^2)

where vivf can be found
1/2 m vivf^2=1/2 mvi^2-mg*496

solver for vivf the final vertical velocity

Now, speed at impace
v= sqrt(vivf^2+vih^2)

angle:
arctan=vivf/vih

Answer 2

I don't understand the angle part. But thank you so much for the rest of it!