Two stunt men are going to jump out of a stationary balloon 457meters off the gr

Question

Two stunt men are going to jump out of a stationary balloon 457meters off the ground, jumper A will jump first, pulling hisparachute at 4 seconds after the jump. Jumper B will jump exactly20 seconds after jumper A. Jumper B is slightly heavier than jumperA, when their chutes are open, jumper A falls at 3.0 m/s whilejumper B falls at 3.2m/s. After how many seconds will jumper B needto pull his parachute after leaving the balloon to land at exactlythe same time as jumper A on the ground? How long does the entirestunt take from the instant jumper A leaves the balloon?

I got the second part quickly, the entire stunt should take around130 seconds. By estimation I can figure jumper B needs to open at 5seconds, but I need help setting this up in an equation and showingproof. Thanks!!!

Explanation / Answer

Velocity of first jumper after 4 second of jump = u+at = 0+ 9.8 * 4= 39.2 m/s distance traveled in that 4 second = ut +at2/2 = 0 +9.8*42/2=78.4 m So remaining height = 457 - 78.4 = 378.6 m velocity after parachute opens = 3.0 m/s Time taken for rest of the fall= distance / velocity= 378.6 / 3= 126.2 s So total time taken = 130.2 sec For second diver , he jumps 20 sec after first diver So he must reach bottom in 110.2 sec so both the divers reachsimultaneously . Let he opens the parachute after t second . So distance covered in those t second = 9.8*t2/2 =4.9t2 So remaining time = 110.2 - t He falls with speed of 3.2 m/s So distance traveled = (110.2-t) *3.2 = 352.64 - 3.2 t but total height = 457 m So 4.9t2 + 352.64 - 3.2t = 457 Solve the equation to get t You will get t=4.95 sec and -4.3 sec But time cannot be negative , so t=4.95 sec Second diver opens the chute after 4.95 second of his fall

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