Question

Question

Answer 1

Answer:

The light bulb will reach the ground 1.25 seconds after it is dropped.

Step-by-step explanation:

We know that for an object that is in the air, the only force acting on it will be the gravitational force (where we are ignoring the air resistance)

Then the acceleration of the object is the gravitational acceleration, 32.17 ft/s^2

Then the acceleration of the light bulb is:

A(t) = (-32.17 ft/s^2)

Where the negative sign is because the acceleration is downwards.

Now, to get the velocity equation, we need to integrate the acceleration over time, we will get:

V(t) = (-32.17 ft/s^2)*t + V0

Where V0 is the initial velocity of the light bulb. Because it is dropped, the initial velocity will be zero, then V0 = 0m/s, then the velocity equation is:

V(t) =  (-32.17 ft/s^2)*t

Finally, to get the position equation we need to integrate again, we will get:

P(t) = (1/2)*(-32.17 ft/s^2)*t^2 + P0

Where P0 is the initial height of the object, and in this case, we know that it is equal to 25 ft.

Then the position equation is:

P(t) = (1/2)*(-32.17 ft/s^2)*t^2 + 25ft

The object will hit the ground when P(t) = 0 ft, then we need to solve that equation for t:

P(t) =  (1/2)*(-32.17 ft/s^2)*t^2 + 25ft = 0 ft

          25 ft =  (1/2)*(32.17 ft/s^2)*t^2

         2*25ft = (32.17 ft/s^2)*t^2

           50ft =  (32.17 ft/s^2)*t^2

         √( 50ft/(32.17 ft/s^2)) = t = 1.25 s

The light bulb will reach the ground 1.25 seconds after it is dropped.