A cannonball is shot from the top of a 47.8 m high hill at the speed of 53.7 m/s. At its highest point, it is moving 31.5 m. How high is it then?
+36916 I come up with 144.2 m IF in your question, 31.5 m is actually 31.5 m/s
+36916 I come up with 144.2 m IF in your question, 31.5 m is actually 31.5 m/s
If , at its HIGHEST point , it is still moving 31.5 m/s that value must be the HORIZONTAL component of the initial velocity (as the vertical component is ZERO when at hightest point.
SO angle of shot is 53.7 cos(angle) = 31.5 Yields angle = 54.08 degrees
Then VERTICAL component is 53.7 sin(54.08) = 43.49 m/s
vf=vo + at 0 = 43.49 + (-9.81)t yields t = 4.433 seconds
xf=xo + vo t + 1/2 a t^2 yields xf = 144.2 m
