An object is thrown up into the air with an initial velocity of 50 m/s a) How hi

Question

An object is thrown up into the air with an initial velocity of 50 m/s a) How high does it go up? b) where is it after 4 seconds with respect to the floor? An object is released from rest from a top of the building 80 meters high where is it (as me from the top of a building) 4 seconds later and what is its speed? An object of mass 5 kg moving with a velocity of 8m/s at a height of 20 meters above the floor what is its kinetic potential and total energies at that instant. An object of mass 20 kg moving to the right with a speed of 5m/s collides with an object motioning towards it having a mass of 5kg and moving with a speed of 4m/s which way is the 20 kg mass is moving and with what speed after the collision

Explanation / Answer

1A.

Vi = 50 m/sec

a = -9.81 m/sec^2

at the top Vf = 0

Using equation

Vf^2 = Vi^2 + 2*a*h

h = (Vf^2 - Vi^2)/2a

h = (0^2 - 50^2)/(-2*9.81) = 127.42 m

B.

at t = 4 sec, using equation:

h1 = Vi*t + 0.5*a*t^2

h1 = 50*4 - 0.5*9.81*4^2 = 121.52 m above the ground

2.

Object is released, So

Vi = initial speed = 0

a = acceleration = 9.81 m/sec^2

t = 4 sec

Using equation:

h = Vi*t + 0.5*a*t^2

h = 0*4 + 0.5*9.81*4^2

h = 78.48 m

object is 78.48 m below the top of building

Vf = Vi + a*t

Vf = 0 + 9.81*4 = 39.24 m/sec

At t = 4 sec speed of object is = 39.24 m/sec

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