The rocket deploys a parachute after 9 seconds and descends at a constant rate of 20 ft/sec. How long does it take to reach the ground

This question is incomplete, the complete question is;

A rocket is launched straight into the air with initial velocity 200 ft/s. Assume the launch is instantaneous. How high will it go, and how long will that take (use time increments of 0.01 seconds)? G= - 32.2 ft/s² , H₀=0 (the ground)

V(t)=V₀ + g t

H(t)= H₀ + V₀ t + ½ g t²

The rocket deploys a parachute after 9 seconds and descends at a constant rate of 20 ft/sec. How long does it take to reach the ground

a) maximum height S is 621.12 ft and time taken to reach it is 6.21 sec.

b) The rocket deploys a parachute after 9 seconds and descends at a constant rate of 20 ft/sec,

Time taken to travel t = 24.79 sec

NOW total time taken = 27.55 sec

Step-by-step explanation:

Given that,

initial velocity u = 200 ft/s

acceleration a = 32.2 ft/s²

vertical displacement ( maximum height) S = ?

final velocity at highest point Ц = 0

NOW from the equation of motion

v² - u² = 2as

we substitute

0² - (200)² = 2 (-32.2) × s

s = 621.12 ft

time taken t = (v -u)/a = (0-200)/-32.2

t = (-200) / (-32.2)

t = 6.21 sec

distance traveled by rocket from t=6.21sec to t=9sec

Δt = (9-6.21) = 2.79sec

s = ut + 1/2at²

s = 0 + 1/2(32.2)(2.79)² = 125.32 ft

Remaining distance = 621.12 ft - 125.32 ft = 495.8 ft

Time taken to travel = t = s/v = (495.5ft) / (20 ft/sec)

t = 24.79 sec

NOW total time taken = 2.79sec + 24.79sec = 27.55 sec

54 km / hours

step-by-step explanation:

we need to convert seconds to hours

60 seconds = 1 minutes

60 minutes = 1hour

15 m         60 second   60 minutes

  * *         = 54000 meters/ hour

second     1 minute             1 hour

now we need to convert meters to km

1000 m = 1 km

54000 m       1 km

*   = 54 km / hours

hour               1000m

$5.41

step-by-step explanation:

$20 - $14.51 = $5.49.