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

Determine:

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

(a) 29.71 m/s

(b) - 882.68 m/s²

(c) 0.0337 s

Explanation:

(a)

Using

v² = u² + 2gh ... Equation 1

Where v = final speed, u = initial speed, g = acceleration due to gravity, h = height.

Given: u = 0 m/s (falling from a height), h = 45 m, g = 9.81 m/s²

Substitute into equation 1

v² = 0² + 2 (9.81) (45)

v² = 882.9

v = √882.9

v = 29.71 m/s.

Hence the speed of the box just before it touches the top of the car = 29.71 m/s.

(b)

Using the formula,

v² = u² + 2as

Where s = depth of crushed due to impact.

make a the subject of the equation

a = (v²-u²) / 2s ... Equation 2

Note: At the time when the box land on the car, v = 0 m/s, and u = 29.71 m/s.

Given: v = 0 m/s, u = 29.71 m/s, s = 0.50 m

Substitute into equation 2

a = (0-29.71²) / (2*0.5)

a = - 882.68 m/s²

Note: a is negative because the box is decelerating.

(c)

using

v = u + at

Where t = time it took to crush the roof.

making t the subject of the equation,

t = (v-u) / a ... Equation 3

Given: v = 0 m/s, u = 29.71, a = - 882.68 m/s²

Substitute into equation 3

t = (0-29.71) / -882.68

t = - 29.72/-882.68

t = 0.0337 s