Physics, 29.10.2019 17:31 gloriahope3017

Arock is thrown horizontally from a cliff at 25 m/s.
a) what will its horizontal and vertical velocities be after 2.0 s?
b) what will its velocity be (magnitude and direction) after 3.0 s?
c) what will its displacement be after 3.0 s?

Answers: 2

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Answer from: galayeet

a) horizontal: 25.0 m/s, vertical: -19.6 m/s

As the rock is thrown horizontally, it follows a parabolic motion (projectile motion), where:

- The horizontal motion is a uniform motion, with constant velocity

- The vertical motion is a uniformly accelerated motion, with constant acceleration g=-9.8 m/s^2 towards the ground (we take upward as positive direction, so the acceleration is negative)

We said that the horizontal velocity is constant, so it remains equal to the velocity at which the rock was thrown from the cliff, therefore

v_x = 25 m/s

And this is also the value of the horizontal velocity at t = 2.0 s.

Instead, the vertical velocity follows the equation

v_y = u_y +gt

where:

u_y = 0 is the initial vertical velocity

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

t = 2.0 s is the time

Substituting,

v_y = 0 +(-9.8)(2.0)=-19.6 m/s

b) 38.6 m/s at $-49.6^{\circ}$

In order to find the velocity at 3.0 s, we should find its horizontal and vertical components at that time first.

As we said, the horizontal velocity is constant:

v_x = 25 m/s

while the vertical velocity is given by

v_y = u_y +gt

And substituting t = 3.0 s,

v_y = 0 +(-9.8)(3.0)=-29.4 m/s

The magnitude of the velocity at 3.0 s therefore is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(25)^2+(-29.4)^2}=38.6 m/s$

While the direction is given by

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-29.4}{25.0})=-49.6^{\circ}$

where the negative sign means the angle is below the horizontal.

c) 87.0 m

In order to find the displacement, we should find the horizontal and vertical distance travelled by the rock in 3.0 s.

Along the horizontal direction, the motion is uniform, so the horizontal distance travelled is

d_x = v_x t = (25)(3.0)=75 m

While the distance travelled along the vertical direction is given by the equation

$d_y = u_y t + \frac{1}{2}gt^2$

And substituting t = 3.0 s,

$d_y = 0 + \frac{1}{2}(-9.8)(3.0)^2=-44.1 m$

So, the displacement of the rock after 3.0 s is

$d=\sqrt{d_x^2+d_y^2}=\sqrt{75^2+(-44.1)^2}=87.0 m$

Answer from: Quest

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Answer from: Quest

i think the answer is a and b

i wish you luck

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Arock is thrown horizontally from a cliff at 25 m/s.
a) what will its horizontal and vertical...