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UkoKoshka [18]
3 years ago
10

An object is thrown horizontally off a cliff with an initial velocity of 5.0 meters per second. The object strikes the ground 3.

0 seconds later. What is the vertical speed of the object as it reaches the ground? [Neglect friction.]

Physics
2 answers:
vladimir2022 [97]3 years ago
6 0

The acceleration of gravity is 9.8 m/s².  This simply means that when anything falls, its downward speed keeps increasing, and it falls 9.8 m/s faster every second than it fell 1 second earlier.

After 3 seconds of falling, the object is falling at (3 x 9.8 m/s) = 29.4 m/s faster than at the beginning of the 3 seconds.  If it had no vertical speed at the beginning of the 3 seconds, then THAT's its speed after 3 seconds . . . . . <em>29.4 m/s</em> downward.

As far as being thrown horizontally off the cliff . . . that has no effect on it vertical speed.  Horizontally, it doesn't matter whether it rolls gently over the edge, or somebody throws it horizontally, or it gets shot horizontally out of a high power rifle.  It hits the ground at the same time and with the same speed in every case.

Zigmanuir [339]3 years ago
3 0

The vertical speed of the object as it reaches the ground is about 29 m/s

\texttt{ }

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration (m / s²)v = final velocity (m / s)</em>

<em>u = initial velocity (m / s)</em>

<em>t = time taken (s)</em>

<em>d = distance (m)</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

initial horizontal speed of the object = u_x = 5.0 m/s

initial vertical speed of the object = u_y = 0 m/s

elapsed time = t = 3.0 s

<u>Unknown:</u>

final vertical speed of the object = v_y = ?

<u>Solution:</u>

<em>The motion of the object is a </em><em>parabolic motion.</em>

<em>We can use the following formula to calculate </em><em>the vertical speed </em><em>of the object as it reaches the ground:</em>

v_y = u_y + gt

v_y = 0 + 9.8( 3.0 )

v_y = 29.4 \texttt{ m/s}

v_y \approx 29 \texttt{ m/s}

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

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A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2
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Answer:

Approximately 0.608 (assuming that g = 9.81\; \rm N\cdot kg^{-1}.)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

  • Let m represent the mass of this car.
  • Let r represent the radius of the circular track.

This answer will approach this question in two steps:

  • Step one: determine the centripetal force when the car is about to skid.
  • Step two: calculate the coefficient of static friction.

For simplicity, let a_{T} represent the tangential acceleration (1.90\; \rm m \cdot s^{-2}) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to 90^\circ or \displaystyle \frac{\pi}{2} radians.

The angular acceleration of this car can be found as \displaystyle \alpha = \frac{a_{T}}{r}. (a_T is the tangential acceleration of the car, and r is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity (u and v) to (tangential) acceleration a_{T} and displacement x:

v^2 - u^2 = 2\, a_{T}\cdot x.

The idea is to solve for the final angular velocity using the angular analogy of that equation:

\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta.

In this equation, \theta represents angular displacement. For this motion in particular:

  • \omega(\text{initial}) = 0 since the car was initially not moving.
  • \theta = \displaystyle \frac{\pi}{2} since the car travelled one-quarter of the circle.

Solve this equation for \omega(\text{final}) in terms of a_T and r:

\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}.

Let m represent the mass of this car. The centripetal force at this moment would be:

\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}.

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, m\, g.

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let \mu_s denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g.

The size of this force should be equal to that of the centripetal force when the car is about to skid:

\mu_s\, m\, g = \pi\, m\, a_{T}.

Solve this equation for \mu_s:

\mu_s = \displaystyle \frac{\pi\, a_T}{g}.

Indeed, the expression for \mu_s does not include any unknown letter. Let g = 9.81\; \rm N\cdot kg^{-1}. Evaluate this expression for a_T = 1.90\;\rm m \cdot s^{-2}:

\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608.

(Three significant figures.)

7 0
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