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a diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. she starts with a velocity of 4.00 m/s, and her takeoff point is 1.90 m above the pool.
(a) what is her highest point (in m) above the board?
_____ m
(b) how long a time (in s) are her feet in the air?
_____ s
(c) what is her velocity (in m/s) when her feet hit the water? (assume the positive direction is upward. indicate the direction with the sign of your answer.)
_____ m/s

Explanation:

Step1: Define known values

Initial velocity $v_0 = 4.00\ \text{m/s}$, acceleration $a = -g = -9.81\ \text{m/s}^2$, takeoff height above pool $y_0 = 1.90\ \text{m}$
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Part (a)

Step1: At highest point, $v=0$

Use $v^2 = v_0^2 + 2a\Delta y$

Step2: Solve for height above board

$\Delta y = \frac{v^2 - v_0^2}{2a} = \frac{0 - (4.00)^2}{2(-9.81)} = \frac{-16}{-19.62} \approx 0.815\ \text{m}$
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Part (b)

Step1: Time to reach highest point

Use $v = v_0 + at_1$, solve $t_1 = \frac{v - v_0}{a} = \frac{0 - 4.00}{-9.81} \approx 0.408\ \text{s}$

Step2: Displacement from peak to pool

Total downward displacement: $\Delta y_{\text{down}} = 0.815 + 1.90 = 2.715\ \text{m}$

Step3: Time to fall from peak

Use $\Delta y_{\text{down}} = \frac{1}{2}gt_2^2$, solve $t_2 = \sqrt{\frac{2\Delta y_{\text{down}}}{g}} = \sqrt{\frac{2(2.715)}{9.81}} \approx 0.745\ \text{s}$

Step4: Total time in air

$t_{\text{total}} = t_1 + t_2 = 0.408 + 0.745 \approx 1.15\ \text{s}$
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Part (c)

Step1: Displacement from board to pool

$\Delta y = -1.90\ \text{m}$ (downward from board)

Step2: Use $v^2 = v_0^2 + 2a\Delta y$

$v = \sqrt{v_0^2 + 2a\Delta y} = \sqrt{(4.00)^2 + 2(-9.81)(-1.90)}$
$= \sqrt{16 + 37.278} = \sqrt{53.278} \approx -7.30\ \text{m/s}$ (negative for downward direction)

Answer:

(a) $0.815\ \text{m}$
(b) $1.15\ \text{s}$
(c) $-7.30\ \text{m/s}$