1. ## Equation of motion help please.

A rock is thrown horizontally from a cliff with a speed of 15 m/s. It falls half the height of the cliff in the last three seconds of its fall. What is the total fall time? (use g = 9.81 m/s^2)

This is my attempt, since the rock is thrown horizontally it makes no difference to the fall time.

so:

$\frac{3}{\left(1-\sqrt1-0.5\right)}=10.24264\:secs$

2. Can any one help please? i'm convinced my time is correct.

$\frac{1}{2}at^2 = 4.905\times7.24264^2 = 257.295\:m$

$\frac{1}{2}at^2 = 4.905\times10.24264^2 = 514.59\:m$

$\frac{257.295}{514.95}= 0.5$

3. Hello, 1LISTEN!

A rock is thrown horizontally from a cliff with a speed of 15 m/s.
It falls half the height of the cliff in the last three seconds of its fall.
What is the total fall time? .(Use g = 9.81 m/sē) .
Also not needed

Since the rock is thrown horizontally its initial speed
makes no difference to the fall time. . . . . . Right!

The height of an dropped object is: . $y(t) \:=\:h_o - \frac{1}{2}gt^2$
. . where $h_o$ = initial height.

In the first $T$ seconds, the rock falls half the height: . $y(T) \,=\,\frac{1}{2}h_o$
. . $h_o - \frac{1}{2}gT^2 \:=\:\frac{1}{2}h_o \quad\Rightarrow\quad h_o \:=\:gT^2$ .[1]

In the next 3 seconds, the rock strikes the ground: . $y(T+3) \,=\,0$
. . $h_o - \frac{1}{2}g(T+3)^2 \:=\:0 \quad\Rightarrow\quad h_o \:=\:\frac{1}{2}g(T+3)^2$ .[2]

Equate [1] and [2]: . $gT^2 \:=\:\frac{1}{2}g(T+3)^2 \quad\Rightarrow\quad T^2 - 6T - 9 \:=\:0$

Quadratic Formula: . $T \;=\;\frac{6\pm\sqrt{72}}{2} \;=\;3 \pm3\sqrt{2}$

Therefore, the total fall time is: . $T + 3 \;=\;6 + 3\sqrt{2}$ seconds.