Question

A construction worker riding an open elevator is moving upward with a speed of 6.0 m/s. When he is 42 m above the ground he accidentally drops a hammer. What are the maximum height reached by the hammer, the time it takes for the hammer to fall to the ground and speed of the hammer when it hits the ground? Please answer all parts, thank you!

Answer 1

When a hammer is dropped from a elevator moving with speed 6 m/s, hammer is moving upwards with speed 6 m/s immediately after dropping. Hence vertical distance h travelled by hammer is calculated form the equation

" v2 = u2 - 2gh " , where v is final speed that equals zero when hammer reaches maximum height,

u = 6 m/s is initial speed and g is acceleration due to gravity.

Hence we get h = u2 / (2g) = ( 6 6 ) / ( 2 9.8 ) = 1.837 m

Time taken by the hammer to hit the ground is calculated from the relation

H = (1/2)gt2

In the above equation , H is vertical height travelled by the hammer from its

maximum height , i.e. ( 42 + 1.837 ) = 43.837 m

time taken to hit the ground = ( 2H / g )1/2 = ( 2 43.837 / 9.8 )1/2 3 s

Speed v of the hammer when it hit the ground is calculated using the relation

v2 = u2 + 2 g H

where u is intial speed which is zero when hammer starts falling from maximum height H and g is acceleration due to gravity .

v = { 2 9.8 43.837 }1/2 = 29.31 m/s