Question

1-D Kinematics, Constant Acceleration After falling a distance of 45.0 m from the top of a building, a box is landing on the top of a car, which is crushed to a depth of 0.50 m due to the impact. g = 9.8m/s2 Calculate: a) the speed of the box just before it touches the top of the car. b) The average acceleration of the box during the impact. c) The time it took to crush the roof.

Answer 1

Answer:

Explanation:

Given

Object fall from a height of $s=45\ m$

Considering initial velocity to be zero i.e. $u=0$

using

$v^2-u^2=2as$  

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-0=2(9.8)\cdot 45$

$v=29.69\ m/s\approx 29.7\ m/s$

(b)Average acceleration

After falling 45 m, object strike the car and comes to rest after covering a distance of 0.5 m

again using

$v'^2-u'^2=2as'$

here final velocity will be zero i.e.$v'=0$

initial velocity $u'=v$

$0-(29.7)^2=2\cdot a\cdot 0.5$

$a=-882.09\ m/s^2$

(c)time taken by it to stop

$v'=u'+a't$

$0=29.7-882.09\cdot t$

$t=0.034\ s$