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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. how far from the base of the cliff

Respuesta :

You know that the formula for finding horizontal displacement is; 

s=uₓt 
s=(5 m/s)(3s) 
s=15 m

Therefore, the ball travelled 15 m. 

Hope I helped :) 

The object will strike the ground at a distance of [tex]\boxed{15\text{ m}}[/tex] from the base of the cliff.

Further explanation:

This is a problem of projectile motion. When an object is thrown horizontally from a certain height, the object moves both in X and Y direction under the action of the acceleration due to gravity.

Given:

The initial velocity with which an object is thrown horizontally is [tex]5.0\text{ m/s}[/tex].

The object strikes the ground [tex]3.0\text{ s}[/tex] later so the total time of flight is [tex]3.0\text{ s}[/tex].

Concept:

First we choose the coordinate axis.

So let’s assume east direction as the positive X axis and vertical upward direction as the positive Y axis.

During the whole flight object is subjected to a downward acceleration [tex]g[/tex]. In horizontal direction external force on the object is zero so acceleration in X direction will be zero.

Analyze the motion of object in both X and Y direction:

In X direction,

[tex]{a_x}=0\\{u_x}=5.0\,{\text{m/s}}[/tex]

Let the distance traveled by an object in X-direction is [tex]{S_{\text{x}}}[/tex].

Acceleration is defined as the rate of change of velocity. Here, in X direction the acceleration is zero; therefore velocity of object will remain same in X direction throughout the motion.

Use the second equation of motion:

[tex]{S_{\text{x}}}={u_{\text{x}}}t+\dfrac{1}{2}{a_{\text{x}}}{t^2}[/tex]

Substitute [tex]5.0\,{\text{m/s}}[/tex] for [tex]{u_x}[/tex], [tex]3\text{ s}[/tex] for [tex]t[/tex] and [tex]0[/tex] for [tex]{a_x}[/tex] in the above expression.

[tex]\begin{aligned}{S_{\text{x}}}&=\left( {5.0\,{\text{m/s}}} \right)\left( {3\,{\text{s}}} \right)+\frac{1}{2}\left( 0 \right){t^2}\\&=15.0\,{\text{m}}\\\end{aligned}[/tex]

Thus, the object will strike the ground at a distance of [tex]\boxed{15\text{ m}}[/tex] from the base of the cliff.

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Projectile motion

Keywords:

object, horizontally, thrown, cliff, height, initial velocity, strikes, ground, base, 5m/s, 3 sec, distance, acceleration due to gravity, X direction, Y-direction.

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