Question

A ball shot at an angle of 60° to the ground strikes a building 23m away at a point 16m
high. Find the magnitude and direction of the final velocity of the ball as it strikes the wall.

Answer 1

Angle of elevation 60
initial speed = s
initial vertical speed Vi
= s sin 60 = .866 s
horizontal speed forever = u = s cos 60
= s/2

t = 23/u = 46/s

at t = 46/s, h = 16

16 = 0 + Vi t - 4.9 t^2
16 = .866 s (46/s) - 4.9*46^2/s^2
-23.8 = -10368/s^2
s^2 = 435.6
s = 20.9 m/s
That is INITIAL speed
now t = 46/s = 2.2 seconds
Vi = .866 * 20.9 = 18.1
u = .5 * 20.9 = 10.5
v = 18.1 - 9.8 (2.2) = - 3.46
magnitude of speed = sqrt(u^2+v^2)
= sqrt (10.5^2 + 3.46^2)
= 11.1 m/s
tan angle down from horizontal = 3.46/10.5
so angle down from horizontal = 18.2 degrees downward