Question

Suppose a car manufacturer tested its cars for front-en4 collisions by hauling them up on a crane and dropping then; from a certain height, (a) Show that the speed just before a car hits the ground, after falling from rest a vertical distance H, is given by \/2 g H . What height corresponds tq a collision at (b) 50 km/h

Answer 1

a

Generally from third equation of motion we have that

`v^2 = u^2 + 2a[s_i - s_f]`

Here v is the final speed of the car

u is the initial speed of the car which is zero

`s_i` is the initial position of the car which is certain height H

`s_i` is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

`v^2 = 0 + 2g[H - 0]`

=>
`v = √( 2 g H)`

b

`H = 9.86 \ m`

Step-by-step explanation:

Generally from third equation of motion we have that

`v^2 = u^2 + 2a[s_i - s_f]`

Here v is the final speed of the car

u is the initial speed of the car which is zero

`s_i` is the initial position of the car which is certain height H

`s_i` is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

`v^2 = 0 + 2g[H - 0]`

=>
`v = √( 2 g H)`

When
`v = 50 \ km/h = (50 *1000)/(3600) = 13.9 \ m/s` we have that

`13.9 = √( 2 g H)`

=>
`H = (13.9^2)/(2 * 9.8)`

=>
`H = 9.86 \ m`