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
Answer:
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
