Answer:
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
