A diver launches off of a 25 m cliff with a speed of 5 m/s and an angle of 30° from the horizontal. How long does it take the diver to hit the water?

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A diver launches off of a 25 m cliff with a speed of 5 m/s and an angle of 30° from the horizontal. How long does it take the diver to hit the water?

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Answer 1 · 2016-02-22T06:44:00

Assuming $30^{o}$ $30^o$ is taken below the horizontal
$t ≅ 2.0$ $t~=2.0$ $s$ $s$ .
Assuming $30^{o}$ $30^o$ is taken above the horizontal
$t ≅ 2.5$ $t~=2.5$ $s$ $s$ .

Explanation:

Once you know the initial velocity in the y, you can treat this as one dimensional motion (in the y) and ignore the x motion (you only need the x if you want to know how far from the cliff they will land).

Note: I will be treating UP as negative and DOWN as positive for the WHOLE problem.

-Need to know if it's $30^{o}$ $30^o$ above or below the horizontal (you probably have a picture)

A) Assuming $30^{o}$ $30^o$ below the horizontal, (she jumps down).

We break up the initial velocity of $5$ $5$ $\frac{m}{s}$ $m/s$ as follows:
$v_{y} = 5 \cdot sin (30^{o})$ $v_y = 5 * sin (30^o)$ $\frac{m}{s} [$ $m / s [$ down $]$ $]$
$v_{y} = 5 \cdot 0.5$ $v_y = 5 * 0.5$ $\frac{m}{s} [$ $m / s [$ down $]$ $]$

$v_{y} = + 2.5$ $v_y = + 2.5$ $\frac{m}{s}$ $m / s$
Note that $v_{x} = 5 \cdot cos (30^{o})$ $v_x = 5 * cos (30^o)$ $\frac{m}{s} [$ $m / s [$ away from cliff $]$ $]$ ,
but this does NOT affect the answer.

We have the initial velocity $v_{1}$ $v_1$ or $v_{o} = 2.5$ $v_o=2.5$ $\frac{m}{s}$ $m/s$ in the y,
the acceleration, $a$ $a$ , in the y (just gravity $a = 9.8$ $a = 9.8$ $\frac{m}{s^{2}}$ $m/s^2$ ),
the displacement, $d = 25$ $d=25$ $m$ $m$ , in the y and want the time, $t$ $t$ .
The Kinematic equation that has these terms is given by:
$d = v_{1} t + \frac{1}{2} a t^{2}$ $d=v_1 t +1/2 at^2$
Subbing in we have
$25$ $25$ $m = 2.5$ $m=2.5$ $\frac{m}{s} t + \frac{1}{2} 9.80$ $m/s t +1/2 9.80$ $\frac{m}{s^{2}}$ $m/s^2$ $t^{2}$ $t^2$ , dropping the units for convince and rearranging we have

$0 = 4.90$ $0=4.90$ $t^{2} + 2.5$ $t^2 + 2.5$ $t - 25$ $t-25$
Put this though the quadratic formula to solve for t.
$t_{1} = - 2.5282315724434$ $t_1=-2.5282315724434$ and
$t_{2} = 2.0180274908108$ $t_2=2.0180274908108$ .
In this case the negative root is nonsense, so $t ≅ 2.0$ $t~=2.0$ $s$ $s$ .

B) Assuming $30^{o}$ $30^o$ above the horizontal, (she jumps up).

We break up the initial velocity of $5$ $5$ $\frac{m}{s}$ $m/s$ as follows:
$v_{y} = 5 \cdot sin (30^{o})$ $v_y = 5 * sin (30^o)$ $\frac{m}{s} [$ $m / s [$ up $]$ $]$

$v_{y} = 5 \cdot 0.5$ $v_y = 5 * 0.5$ $\frac{m}{s} [$ $m / s [$ up $]$ $]$

$v_{y} = - 2.5$ $v_y = - 2.5$ $\frac{m}{s}$ $m / s$ (positive is down and negative is up!)
Note that $v_{x} = 5 \cdot cos (30^{o})$ $v_x = 5 * cos (30^o)$ $\frac{m}{s} [$ $m / s [$ away from cliff $]$ $]$ , but this does NOT affect the answer.

We have the initial velocity $v_{1}$ $v_1$ or $v_{o} = - 2.5$ $v_o=-2.5$ $\frac{m}{s}$ $m/s$ in the y, the acceleration, $a$ $a$ , in the y (just gravity $a = 9.8$ $a = 9.8$ $\frac{m}{s^{2}}$ $m/s^2$ ), the displacement, $d = 25$ $d=25$ $m$ $m$ , in the y and want the time, $t$ $t$ . The Kinematic equation that has these terms is given by:
$d = v_{1} t + \frac{1}{2} a t^{2}$ $d=v_1 t +1/2 at^2$
Subbing in we have
$25$ $25$ $m = - 2.5$ $m=-2.5$ $\frac{m}{s} t + \frac{1}{2} 9.80$ $m/s t +1/2 9.80$ $\frac{m}{s^{2}}$ $m/s^2$ $t^{2}$ $t^2$ , dropping the units for convince and rearranging we have

$0 = 4.90$ $0=4.90$ $t^{2} - 2.5$ $t^2 - 2.5$ $t - 25$ $t-25$
Put this though the quadratic formula to solve for t.
$t_{1} = - 2.0180274908108$ $t_1=-2.0180274908108$ and
$t_{2} = 2.5282315724434$ $t_2=2.5282315724434$ (look they switched!)

Again, the negative root is nonsense, so $t ≅ 2.5$ $t~=2.5$ $s$ $s$ .

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A diver launches off of a 25 m cliff with a speed of 5 m/s and an angle of 30° from the horizontal. How long does it take the diver to hit the water?

1 Answer

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