A particle falling under gravity describes 80 ft in a certain seconds. How long does it take to describe the next 112ft? [$ g = 32ft/{sec^2}$]
A) 1s
B) 2s
C) 3s
D) 4s

Hint: Here the particle is falling under the influence of gravity. Here we would apply the equations of gravity and find out the height of the particle from the ground. Then apply the second equation of kinematics in which distance initial velocity, acceleration and time is involved.

Formula used:
${v^2} = {u^2} + 2gh$;
Where:
v = final velocity;
u = initial velocity;
g = gravitational acceleration;
h = height;
$s = vt + \dfrac{1}{2}a{t^2}$;
Where:
s = distance;
v = initial velocity;
a = g = Gravitational acceleration
t = time;

Complete step by step solution:
Apply the equation of kinematics and solve for the unknown time t.
${v^2} = {u^2} + 2gh$;
The initial velocity is zero;
 ${v^2} = 0 + 2gh$;
Put in the rest of the given values in the above equation:
${v^2} = 2 \times 32 \times 80$;
Solve mathematically;
${v^2} = 5120$;
To remove the square on the LHS put square root on the RHS:
$v = \sqrt {5120} $;
$v = 71.55$ft/s.
Now this velocity is the initial velocity for the 112ft.
Apply the equation of kinematics:
$s = vt + \dfrac{1}{2}a{t^2}$;
Put in the given values:
$112 = 71.55t + \dfrac{1}{2} \times 32{t^2}$;
Simplify the above equation:
$0 = 71.55t + 16{t^2} - 112$;
Factorize the above equation by the formula:
\[t = - b \pm \sqrt {{b^2} - 4ac} /2a\];
Put the value of h in the above equation and solve:
\[t = \dfrac{{ - 16 \pm \sqrt {{{16}^2} - 4 \times 71.55 \times - 112} }}{{2 \times 71.55}}\];
Do the needed mathematical calculations:
\[t = \dfrac{{ - 16 \pm \sqrt {256 - 32054} }}{{143.1}}\];
\[t = \dfrac{{ - 16 \pm \sqrt {31798} }}{{143.1}}\];
Do the subtraction and then divide
\[t = \dfrac{{ - 16 \pm 178.31}}{{143.1}}\];
\[t = \dfrac{{162}}{{143.1}}\]
After solving the above equation, the value comes out to be:
\[t \simeq 1s\].

Hence, Option (A) is correct i.e. It takes 1seconds to describe the next 112ft.

Note: Here for the first case the initial velocity would be zero, apply the equation of kinematics ${v^2} = {u^2} + 2gh$ and find the final velocity v. The final velocity v would be the initial velocity for the second case when the particle is falling and has covered a distance of 112ft. Apply the equation $s = vt + \dfrac{1}{2}a{t^2}$ and solve for time.