An object, initially at rest, is dropped off a building and accelerates to earth at -9.81m/s2. How long will it take for the object to reach a...

For an object falling under the gravity only, we can use the following equation of motion to solve this question:

v = u + at

here, u is the initial velocity of motion, v is the final velocity, a is the acceleration and t is time taken. Here, the initial velocity is zero (that is, u = 0 m/s), since the object was initially at rest. The object is falling down, under the influence of...

For an object falling under the gravity only, we can use the following equation of motion to solve this question:

v = u + at

here, u is the initial velocity of motion, v is the final velocity, a is the acceleration and t is time taken. Here, the initial velocity is zero (that is, u = 0 m/s), since the object was initially at rest. The object is falling down, under the influence of gravity and hence it's acceleration will be equal to the acceleration due to gravity (g), that is, a = -9.81 m/s^2 (negative sign d object falling down). The final velocity, v, is given as -49.1 m/s (negative sign d object falling down).

Thus, -49.1 = 0 + (-9.81)t

or, t = -49.1 / -9.8 = 5 seconds.

Thus, it will take the object 5 seconds to reach the velocity of -49.1 m/s.

Just a note, an object falling through the air will experience air resistance and will ultimately reach a terminal velocity. The object has to have a significant mass to reach the final velocity given in the question. Please see the attached link for more information on terminal velocity.

Hope this helps.