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

Answer:

 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

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$, 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 = $0*1+\frac{1}{2}*9.8*1^2=4.9m$

 Displacement after 2 seconds = $0*2+\frac{1}{2}*9.8*2^2=19.6m$

 Displacement after 3 seconds = $0*3+\frac{1}{2}*9.8*3^2=44.1m$

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