Answer :
The time at which the ball remains in the air after it is thrown vertically upward with an initial velocity of 61 feet per second from an initial height of 12 feet is 4 seconds.
Motion that is normal to a defined horizontal surface is often referred to as vertical motion. The height of a ball launched from a certain height while maintaining a constant speed is modeled as h(t) = at²+bt+c where h(t) is the height of the ball, t is time, and a, b, and c are constants.
At the initial height h = 12 feet and t = 0,
12 = a(0)²+b(0)+c
c = 12
Differentiating equation h(t) = at²+bt+c concerning t, we get, [tex]\frac{dh(t)}{dt}&=2at+b[/tex]
Substituting t = 0 and dh(t)/dt = 61, we get,
[tex]\begin{aligned}61&=2a(0)+b\\b&=61\end{aligned}[/tex]
Again differentiating [tex]\frac{dh(t)}{dt}&=2at+b[/tex] concerning t, we get, [tex]\frac{d^2h(t)}{dt^2}=2a[/tex]
The acceleration caused by gravity on Earth is 32 feet per second. Since the ball travels downward, the value becomes negative. Substituting this in the above equation.
[tex]\begin{aligned}-32&=2a\\a&=-16\end{aligned}[/tex]
Substituting values of a, b, and c in h(t) = at²+bt+c, we get,
h(t) = -16t²+61t+12
Substituting h = 0 in the above equation, the time the ball remains in the air will be,
0 = -16t²+61t+12
0 = 16t²-61t-12
Using the quadratic formula,
[tex]\begin{aligned}t&=\frac{-(-61)\pm\sqrt{(-61)^2-4\times16(-12)}}{2\times16}\\t&=\frac{61\pm67}{32}\\t&=\frac{61+67}{32}\;\text{or}\;\frac{61-67}{32}\\t&=4\;\text{or}\;-\frac{3}{16}\end{aligned}[/tex]
Time cannot be negative, therefore, t = 4 seconds is the answer.
To know more about the vertical motion of the ball:
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