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.
+128406 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
