Question

A gumdrop is released from rest at the top of the empire state building, which is 381 m tall. disregarding air resistance, calculate the displacement and velocity of the gumdrop after 1.00, 2.00, and 3.00 s.

Answer 1

Displacement after 1 second = 4.9 m

Displacement after 2 seconds = 19.6 m

Displacement after 3 seconds = 44.1 m

Velocity after 1 second = 9.8 m/s

Velocity after 2 seconds = 19.6 m/s

Velocity after 3 seconds = 29.4 m/s

Step-by-step explanation:

We have equation of motion ,
`s= ut+(1)/(2) at^2`, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Height of building, displacement = 381 meter

Initial velocity = 0 m/s

Acceleration = Acceleration due to gravity = 9.8
`m/s^2`

Displacement after 1 second =
`0*1+(1)/(2)*9.8*1^2=4.9m`

Displacement after 2 seconds =
`0*2+(1)/(2)*9.8*2^2=19.6m`

Displacement after 3 seconds =
`0*3+(1)/(2)*9.8*3^2=44.1m`

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Velocity after 1 second = 0 + 9.8 * 1 = 9.8 m/s

Velocity after 2 seconds = 0 + 9.8 * 2 = 19.6 m/s

Velocity after 3 seconds = 0 + 9.8 * 3 = 29.4 m/s