Question

An acrobat is launched from a cannon at an angle of 60 degrees above the horizontal. The acrobat is caught by a safety net mounted horizontally at the height from which he was initially launched. Suppose the acrobat is launched at a speed of 26 m/s

a) how long does it take before he reaches his maximum height?

b) How long does it take in total for him to reach a point halfway back down to the ground?

Please help me :) . I have a test tomorrow :)

Answer 1

Vo = 26m/s[60o]
Yo = 26*sin60 = 22.5 m/s.

a. Y = Yo + g*Tr = 0 @ max. ht.
Tr = -Yo/g = -22.5/-9.8 = 2.30 s. = Rise
time.

b. h max = -(Yo^2)/2g = -(22.5)^2/-19.6 = 25.83 m.

h = 0.5g*t^2 = 25.83/2 = 12.91 m.
4.9t^2 = 12.91
t^2 = 2.64
Tf = 1.62 = Fall time.

Tr+Tf = 2.30 + 1.62 = 3.92 s. To fall
halfway back to gnd.