Answer:
(a) 29.71 m/s
(b) -882.68 m/s²
(c) 0.0337 s
Step-by-step 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