A skydiver jumps from an airplane at an altitude of 2,500 ft. He falls under the force of gravity until he opens his parachute at an altitude of 1,000 ft. Approximately how long does the jumper fall before he opens his chute? For this quadratic model we will let the y-axis be the axis of symmetry. h(t) = -16t2 + vt + h0

9.7 seconds

Step-by-step explanation:

From the question, we know that the the skydiver falls under the force of gravity which means His initial velocity (U)= 0

We can calculate His displacement when he's under the influence of gravity by substracting the altitude when he opens his parachute from the altitude he jump with as

S= (2500 ft - 1000 ft)

= 1500 ft

The time taken can be gotten from the 2nd equation of motion which can be expressed mathematically as

Using second equation of motion:

S = u t + 0.5 a t²

Where a= 32ft/s^2 which is the acceleration due to gravity in feet.

S= displacement= 1500 ft

U= initial velocity when under gravity= 0

If all these values is substituted into the Newton equation, we have,

S = u t + 0.5 at²

S= (0×t) + at²

S= 0+at²

t²= S/a

t²= 1500/32

t²= 46.875

t= √46.875

t= 9.67 seconds

Therefore, it will take the the jumper approximately 9.7 seconds to fall before he opens his chute.

9.7 seconds

Step-by-step explanation:

Just did the test, trust me

d

step-by-step explanation:

in a an geometric explicit equation, the first number is the starting rate. in this case 10.