A student drops an object from a fourth floor window. The velocity of the object

is measured at 29.28 m/s 3.0 seconds later. What is the acceleration of the object?

Well, I'm no physicist, but I'll give it a shot! If a student drops an object from a fourth-floor window, the acceleration of the object due to gravity should be roughly 9.8 m/s² downwards. However, be careful with this experiment as the acceleration of students dropping objects may vary depending on their level of enthusiasm or caffeine consumption! It's always good to keep things lighthearted, even when it comes to physics!

To find the acceleration of the object, we can use the kinematic equation:

v = u + at

Where:
v = final velocity of the object
u = initial velocity of the object
a = acceleration
t = time

In this case, the final velocity (v) is given as 29.28 m/s, the initial velocity (u) is assumed to be 0 m/s (since the object was dropped), and the time (t) is given as 3.0 seconds.

Substituting the given values into the equation, we have:

29.28 = 0 + a * 3.0

To isolate the acceleration (a), we can rearrange the equation and solve for a:

a = (29.28 - 0) / 3.0

a = 29.28 / 3.0

a ≈ 9.76 m/s^2

Therefore, the acceleration of the object is approximately 9.76 m/s^2.

To find the acceleration of the object, we can use the equation:

v = u + at

where:
v = final velocity
u = initial velocity
a = acceleration
t = time

In this case, the final velocity (v) is given as 29.28 m/s, the time (t) is 3.0 seconds, and we need to find the acceleration (a).

Since the object is dropped from rest, its initial velocity (u) is 0 m/s.

Substituting the values into the equation:

29.28 = 0 + a * 3.0

To solve for a, we can rearrange the equation:

a = (29.28 - 0) / 3.0

a = 9.76 m/s²

Therefore, the acceleration of the object is 9.76 m/s².

Use v=u+at.

29.28=0+a(3)
a=9.76 m/s^2