Physics, 14.10.2019 04:10 dcarranza626

Aperson standing at the edge of a seaside cliff kicks a stone, horizontally over the edge with a velocity of 18 m/s. the cliff is 52 m above the waters surface
a. how long does it take for the stone to fall to the water
b. what is the vertical velocity component of the stone just before it hits the water
c. what is the horizontal velocity component of the stone just before it hits the water
d. what is the total velocity of the stone just before it hits the water ?

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Answer from: Birlem101

A) 3.26 s

To find the time of flight, we can just analyze the vertical motion of the stone. The vertical displacement is given by

$y=u_y t - \frac{1}{2}gt^2$

where

u_y is the initial vertical velocity

t is the time

g=9.8 m/s^2 is the acceleration of gravity

In this problem we have

y = -52 m (vertical displacement)

u_y=0 (the stone is kicked horizontally, so it has no initial vertical velocity)

Solving for t, we find the time of flight:

$y=-\frac{1}{2}gt^2\\t=\sqrt{\frac{-2y}{g}}=\sqrt{\frac{-2(-52)}{9.8}}=3.26 s$

B) -31.9 m/s

The vertical velocity component at time t is given by the equation

v_y = u_y -gt

where

u_y is the initial vertical velocity

t is the time

g=9.8 m/s^2 is the acceleration of gravity

As we said previously,

u_y = 0

So if we substitute t = 3.26 s (time of flight), we find the vertical velocity when the stone reaches the water:

v_y = 0 - (9.8)(3.26)=-31.9 m/s

where the negative sign means the direction is downward.

C) 18 m/s

In a projectile motion, the horizontal motion is a uniform motion: since there are no forces acting on the projectile along the horizontal direction, the horizontal acceleration is zero, and so the horizontal velocity is constant.

The stone has been initially kicked with an horizontal velocity of

v_x = 18 m/s

So the horizontal velocity just before it hits the water is still 18 m/s.

D) 36.6 m/s at $-60.6^{\circ}$

The total velocity of the stone just before it hits the water is given by:

$v=\sqrt{v_x^2+v_y^2}$

where previously we found that

$v_x = 18 m/s\\v_y = -31.9 m/s$

Substituting,

$v=\sqrt{18^2+(-31.9)^2}=36.6 m/s$

And the direction is

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-31.9}{18})=-60.6^{\circ}$

Answer from: Quest

Answer: it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.explanation: we have equation of motion $s=ut+\frac{1}{2} at^2$ ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/s $s=0*t+\frac{1}{2} *2.18*t^2$ $s=1.09t^2[/ 2equating equation 1 and 2 [tex]4.75t = 1.09t^2\\ \\ t=4.36 seconds.$ so it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.

Answer from: Quest

it's when something is going so fast it makes a loud noise, breaking the sound barrier, for example: a jet makes a loud noise, not only cause of the jet but because they reached a point where they went so fast that you can year the noise from various parts, kinda like a police car, lets say the car was going so fast that you hear the siren of the cop car right next to you, but the car is probably a mile a way, it's cos the siren was next to you a few seconds if you understand me

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