At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40 degrees from horizontal. She leaves the cannon 1m off the ground and lands on the way down in a net 2m off the ground. Determine her horizontal displacement.

Answer 1

Vo = 15 m/s @ 40 Deg.
Xo = 15*cos40 = 11.49 m/s.
Yo = 15*sin40 = 9.64 m/s.

Tr = (Yf-Yo)/g,
Tr = (0-9.64) / -9.8 = 0.984 s. = Time
to rise to max. ht.

hmax = ho + (Yf^2-Yo^2)/2g,
hmax = 1 + (0-(9.64)^2) / -19.6=5.74 m
above gnd.

d = Yo*t + 0.5g*t^2 = 5.74 - 2,
0 + 4.9t^2 = 3.74 m,
t^2 = 0.763,
Tf = 0.874 m. = Time to fall to 2 m
above gnd.

Dx = Xo * (Tr + Tf),
Dx=11.49m/s *(0.984+0.874)s =21.4 m. =
Hor. Displacement.