Question

A ball is thrown upward from the top of a 192-foot-high building. The ball is 240 feet above ground level after 1 second, and it reaches ground level in 6 seconds. The height above ground is a quadratic function of the time after the ball is thrown. Write the equation of this function.

Answer 1

Answer:

The equation of this function is $h = -16\cdot t^{2}+64\cdot t+192$.

Step-by-step explanation:

From Mechanical Physics we must remember that height reached by the ball as a function of time is a quadratic function if and only if ball is uniformly accelerated by gravity and viscous forces and Earth's rotation effects can be neglected. After assuming those considerations as true, the standard form of the quadratic function is:

$h = a\cdot t^{2}+b\cdot t + c$ (Eq. 1)

Where:

$t$ - Time, measured in seconds.

$h$ - Height of the ball, measured in feet.

$a$ - Second order coefficient, measured in feet per square second.

$b$ - First order coefficient, measured in feet per second.

$c$ - Zero order coefficient, measured in feet.

From Algebra, we can obtain a quadratic function by know three distinct points and solving the resulting system of linear equations:

$A(t, h)= (0\,s,192\,ft)$

$c = 192\,ft$ (Eq. 2)

$B(t,h) = (1\,s,240\,ft)$

$a+b+c = 240\,ft$ (Eq. 3)

$C(t,h) =(6\,s,0\,ft)$

$36\cdot a + 6\cdot b + c = 0$ (Eq. 4)

By Algebraic methods, we get this unique solution:

$a = -16\,\frac{ft}{s^{2}}$, $b = 64\,\frac{ft}{s}$, $c = 192\,ft$

The equation of this function is $h = -16\cdot t^{2}+64\cdot t+192$.