Physics Question

Given, four cases.

We know that free-fall motion is independent of mass. 

Finding the amount of time t it takes to hit the ground.

(1) Using the kinematic equation to find t. Downward direction is taken as the negative y-axis:

Symbols used have their usual meanings.

$y-y_0=v_{0y}t-\frac{1}{2}g{t}^2$

Substituting the known values, the initial velocity is zero.

$\begin{array}{rcl}-10-0& =& \left(0\right)t-\frac{1}{2}\left(9.81\right)t^2\\ t^2& =& \frac{2\times 10}{9.81}\\ t& =& \sqrt{\frac{2\times 10}{9.81}}\\ t& =& 1.43 \mathrm{s}\end{array}$

(2) Since, there is no x-component of initial velocity, we will take $v_{0y}=10 \frac{\mathrm{m}}{\mathrm{s}}$

$y-y_0=v_{0y}t-\frac{1}{2}g{t}^2$

Substituting the known values:

$$\begin{gathered} -10-0=\left(10\right)t-\frac{1}{2}\left(9.81\right)t^2 \\ 4.9t^2-10t-10=0 \\ \mathrm{Solving} \mathrm{this} \mathrm{quadratic} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ t=\frac{-\left(-10\right)\pm \sqrt{{\left(10\right)}^2-4\left(4.9\right)\left(-10\right)}}{2\left(4.9\right)} \\ t=\frac{10\pm 17.2}{9.8} \\ \mathrm{Taking} \mathrm{the} \mathrm{positive} \mathrm{value} \mathrm{of} \mathrm{time} \mathrm{as} \mathrm{negative} \mathrm{time} \mathrm{do} \mathrm{not} \mathrm{make} \mathrm{any} \mathrm{sense}: \\ \mathrm{t}=\frac{10+17.2}{9.8} \\ \mathrm{t}=\frac{27.2}{9.8} \\ \mathrm{t}=2.77 \mathrm{s} \end{gathered}$$

(3) Since, there is no x-component of initial velocity, we will take $v_{0y}=10 \frac{\mathrm{m}}{\mathrm{s}}$

$$\begin{gathered} -10-0=\left(10\right)t-\frac{1}{2}\left(9.81\right)t^2 \\ 4.9t^2-10t-10=0 \\ \mathrm{Solving} \mathrm{this} \mathrm{quadratic} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ t=\frac{-\left(-10\right)\pm \sqrt{{\left(10\right)}^2-4\left(4.9\right)\left(-10\right)}}{2\left(4.9\right)} \\ t=\frac{10\pm 17.2}{9.8} \\ \mathrm{Taking} \mathrm{the} \mathrm{positive} \mathrm{value} \mathrm{of} \mathrm{time} \mathrm{as} \mathrm{negative} \mathrm{time} \mathrm{do} \mathrm{not} \mathrm{make} \mathrm{any} \mathrm{sense}: \\ \mathrm{t}=\frac{10+17.2}{9.8} \\ \mathrm{t}=\frac{27.2}{9.8} \\ \mathrm{t}=2.77 \mathrm{s} \end{gathered}$$

The answer will be the same as part (2)

$t=2.77 \mathrm{s}$

(4) Since, there is no x-component of initial velocity, we will take $v_{0y}=10 \frac{\mathrm{m}}{\mathrm{s}}$

$$\begin{gathered} -15-0=\left(10\right)t-\frac{1}{2}\left(9.81\right)t^2 \\ 4.9t^2-10t-15=0 \\ \mathrm{Solving} \mathrm{this} \mathrm{quadratic} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ t=\frac{-\left(-10\right)\pm \sqrt{{\left(10\right)}^2-4\left(4.9\right)\left(-15\right)}}{2\left(4.9\right)} \\ t=\frac{10\pm 19.8}{9.8} \\ \mathrm{Taking} \mathrm{the} \mathrm{positive} \mathrm{value} \mathrm{of} \mathrm{time} \mathrm{as} \mathrm{negative} \mathrm{time} \mathrm{do} \mathrm{not} \mathrm{make} \mathrm{any} \mathrm{sense}: \\ \mathrm{t}=\frac{10+19.8}{9.8} \\ \mathrm{t}=\frac{29.8}{9.8} \\ \mathrm{t}=3.04 \mathrm{s} \end{gathered}$$

So, ranking them 4>(3=2)>1

Answer: last option.