Question

A diver running 1.6 m/s dives out horizontally from the edge of a vertical cliff and reaches the water 3.0 s later. How high was the cliff? How far from its base did the diver hit the water? QUICKLY PLEASE!!!!!!!!!!



h = 47m & d = 15.7m



h= 44m & d = 4.8 m



h = 15.68m & d= 4.8m

Answer 1

Answer:

h= 44m & d = 4.8 m

Explanation:

We're talking about motion in two directions here so make sure you keep track of which direction we're talking about for each question part. Start with your givens, identify your unknown, and find the equation that lets you use the givens to solve for the unknown,.

The first question asks how high the cliff was. This is the vertical direction. We are told the time spent traveling (vertically and horizontally) is 3.0s. Implied is that the vertical acceleration is g, and the initial vertical velocity is 0 m/s. The height of the cliff (vertical distance, y) is unknown. Therefore we are given:

$a_y=g=9.8\\v_0_y=0\\t_y=3\\y=?$

The equation that combines all these is:

$y= v_0_yt_y+ \frac{1}{2}a_yt_y^2\\y=0(3) + \frac{1}{2}(9.8)(3)^2\\y=44.1$

The height of the cliff is 44.1m.

Part 2 asks how far from the cliff base the diver hit the water. This the horizontal direction. We are told the time spent traveling (vertically and horizontally) is 3.0s, and that the horizontal speed is 1.6m/s. The distance from the cliff (horizontal distance, x) is unknown. Therefore we are given:

$v= \frac{\Delta x}{\Delta t}\\1.6= \frac{\Delta x}{3}\\\Delta x =4.8$

The diver hit the water 4.8m from the base.

The correct answer is the second one.

Answer 2

The height of the cliff and horizontal displacement of the diver is required.

The correct option is h= 44 m & d = 4.8 m

$u_x$ = Initial horizontal velocity = 1.6 m/s

$u_y$ = Initial vertical velocity = 0

t = Time taken = 3 s

$a_x$ = Acceleration in x direction = 0

g = $a_y$ = Vertical acceleration = $9.81\ \text{m/s}^2$

Displacement in y direction

$s_y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow s_y=0\times 3+\dfrac{1}{2}\times 9.81\times 3^2\\\Rightarrow s_y=44.145\ \text{m}=h$

Displacement in x direction

$s_x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow s_x=1.6\times 3+\dfrac{1}{2}\times 0\times t^2\\\Rightarrow s_x=4.8\ \text{m}=d$

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