Question

Helppppppppp plzzzzzzzz A parachutist jumps from an airplane and immediately opens her parachute. Her altitude, h, in metres after t seconds is modelled by the equation h(t)=-4t+300. A second parachutist jumps 5 seconds after the first parachutist jumps. Her altitude, in metres, during this time is modelled by the equation h(t)=-4.9^2+49t+177.5. What are both parachutists at the same height when falling? What is the height?

Answer 1

Answer:

Let Ts be the time measured immediately at the jump of the second

parachutist. Then after time T,

the height of the1st parachutist is

y1 = -4(T+5)+300--------(1)

The height of the 2 nd parachutist is

y2=-4.9T^2+300---------(2)

They meets=>

y1=y2=>

-4(T+5)+300=-4.9T^2+300=>

4.9T^2-4T-20=0=>

T=[4+/-sqr(16+4(4.9)(20)]/9.8=>

T=2.469287 or T= -1.65296 (rejected)

Ans. By about 2.47 s after the jump of the 2nd

parachutist, or 7.47 s after the jump of the 1st

parachutist, the 2 parachutists meet in air