Question

119) You throw a rock horizontally off a cliff with a speed of 20 m/s and no significant air resistance. After 2.0 s, the magnitude of the velocity of the rock is closest to

Answer 1

The magnitude of the rock's velocity after 2.0 seconds is 19.6 m/s.

Step-by-step explanation:

The magnitude of the velocity of the rock can be determined using the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the rock is thrown horizontally, so the initial vertical velocity is 0 m/s. The acceleration is due to gravity and is approximately 9.8 m/s². The time is given as 2.0 s. Plugging in these values, the equation becomes:

v = 0 + (9.8)(2.0)

Therefore, the magnitude of the velocity of the rock after 2.0 s is 19.6 m/s.

Answer 2

Answer: 44.54 m/s

Step-by-step explanation:

Let's break the problem into horizontal and vertical.

Horizontal:

Here we do not have any force, only a constant horizontal velocity, so we can write:

Vx(t) = 20m/s.

Vertical:

Here we have the gravitational acceleration acting on the rock, then we can write:

Ay(t) = -9.8m/s^2

For the vertical velocity, we need to integrate over time and get:

Vy(t) = (-9.8m/s^2)*t + V0

Where V0 is the initial vertical velocity, but the rock was thrown horizontally, so we do not have an initial velocity in the vertical axis.

Vy(t) = (-9.8m/s^2)*t.

Ok, now we want to know the magnitude of the velocity when t = 2.0s.

The magnitude will be equal to:

V = √( Vx(2s)^2 + Vy(2s)^2)

= √( (20m/s)^2 + (-2s*9.8m/s^2)^2) = 44.54 m/s

The magnitude of the velocity after 2 seconds is around 44.54 m/s