An object is dropped from 43 feet below the tip of the pinnacle atop a 527-ft tall building. The height h of the object after t
seconds is given by the equation h equals negative 16 t squared plus 484. Find how many seconds pass before the object reaches the ground.
2 answers:
Answer:
5.5 seconds
Step-by-step explanation:
The function that gives the height of the object after t seconds falling is:
h = -16t^2 + 484.
To find the time when the object reaches the ground, we just need to use the value of h = 0 in the equation, and then find the value of t:
0 = -16t^2 + 484.
16t^2 = 484
t^2 = 30.25
t = 5.5 seconds
So the object will reach the ground after 5.5 seconds
Answer:
5.5 sec
Step-by-step explanation:
The height h of the object after t seconds is given by the equation
h =- 16 t²+ 484
The height would be zero when the object hits the ground. therefore equating the above equation to zero
0 = - 16 t²+ 484 (solving for t)
16 t²= 484
t² = 484/16
t²= 30.25
taking square root on B.S
t= 5.5 sec
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