a student drops an object out of the window of the top floor of a high rise dormitory, how fast is the object traveling when it strikes the ground at the end of 3.4s

V = Vo + g*t.

Vo = 0.
g = 9.8 m/s^2.
t = 3.4 s.
V = ?

To determine the speed at which the object is traveling when it strikes the ground, we need to consider the acceleration due to gravity. The acceleration due to gravity is approximately 9.8 meters per second squared (m/s^2), and we can assume it is constant throughout the object's fall.

To find the speed at which the object strikes the ground after 3.4 seconds, we can use the formula for the distance traveled by a falling object:

s = ut + (1/2)at^2

Where:
- s is the distance traveled
- u is the initial velocity (which is 0 since it was dropped)
- a is the acceleration due to gravity
- t is the time elapsed

In this case, we want to solve for the speed at the end of 3.4 seconds, so we need to find the total distance traveled by the object in that time.

s = (1/2)at^2

Plugging in the values:
- a = 9.8 m/s^2
- t = 3.4 s

s = (1/2) * 9.8 * (3.4)^2

s ≈ 56.24 meters

Now that we know the distance traveled, we can calculate the speed.

To calculate the speed, we can use the equation:

v = u + at

Since the initial velocity (u) is 0 (the object was dropped), the equation simplifies to:

v = at

Plugging in the values:
- a = 9.8 m/s^2
- t = 3.4 s

v = 9.8 * 3.4

v ≈ 33.32 meters per second (m/s)

Therefore, the object will be traveling at approximately 33.32 m/s when it strikes the ground after 3.4 seconds.