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A stuntman jumped from an elevation of 500 m and was in free fall before he opened his parachute. His elevation, h meters, t seconds after jumping is modeled by these equations:

h=-4.9t2+2t+500 before the parachute was opened

h=-5t+188 after the parachute was open

a)   How many seconds after the stuntman jumped did he open the parachute? (1 mark)

b)   What was his elevation when he opened the parachute? (1 mark)

Explain your strategy (1 mark). Give your answers to the nearest tenth of a unit.

 Jul 16, 2021
 #1
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b)The  second  equation  gives  his  height when opening the  chute =   188  m

 

a)   The  time from jumping to opening the  chute  can be found as

 

188  = -4.9t^2   +  2t  + 500

 

4.9t^2  - 2t   +  188  - 500   = 0

 

4.9t^2  -  2t  - 312   =  0

 

The time =         ( 2 + sqrt (4  + 4 * 4.9 *312) ) /  (2* 4.9)   ≈  8.186  sec  ≈  8.2 sec

 

 

cool cool cool

 Jul 16, 2021

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