If an object is propelled upward with an initial velocity of 120120 feet per second, then its height in feet after t seconds is given by hequals=120120tminus−16t squaredt2. After how many seconds will the object reach its maximum height? Round your answer to the nearest tenth.

4.3seconds or 3.2seconds to nearest tenth

Step-by-step explanation:

An object propelled upwards with initial velocity of 120ft/s has height modelled by the equation;

H = 120t-26t²

According to equation of motion

V² = U² + 2gH where;

V is the velocity of the object at the maximum height

U is the initial velocity

g is the acceleration due to gravity

H is the maximum height reached

Since the object is propelled upwards, the acceleration due to gravity will be negative. Our equation will now become;

V² = U²-2gH

Given U = 120ft/s

g = 10m/s

Since 1m/s = 3.281ft/s

10m/s = 32.81ft/s

V = 0ft/s(velocity of object at maximum height)

0² = 120²-2(32.8)H

-120² = -65.6H

H = 120²/65.6

H = 219.5ft

To get the time taken by the object to reach the maximum height, we will use

H = 120t - 16t²

219.5 = 120t-16t²

16t²-120t+219.5 = 0

t = -b±√b²-4ac/2a where;

a = 16, b = -120, c = 219.5

t = 120±√120²-4(16)(219.5)/2(16)

t = 120±√14400-14048/32

t = 120±√352/32

t = 120+√352/32 or 120-√352/32

t = 120+18.76/32 or 120-18.76/32

t = 4.3s or 3.2s

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.

not sure but it may be b, as i said i'm not sure.