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

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-09-20T05:22:49+02:00

Answer:

a

Generally from third equation of motion we have that

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

Here v is the final speed of the car

u is the initial speed of the car which is zero

[tex] s_i[/tex] is the initial position of the car which is certain height H

[tex] s_i[/tex] 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

[tex]v^2 = 0 + 2g[H - 0] [/tex]

=> [tex]v = \sqrt{ 2 g H}[/tex]

b

[tex]H = 9.86 \ m [/tex]

Explanation:

Generally from third equation of motion we have that

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

Here v is the final speed of the car

u is the initial speed of the car which is zero

[tex] s_i[/tex] is the initial position of the car which is certain height H

[tex] s_i[/tex] 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

[tex]v^2 = 0 + 2g[H - 0] [/tex]

=> [tex]v = \sqrt{ 2 g H}[/tex]

When [tex]v = 50 \ km/h = \frac{50 *1000}{3600} = 13.9 \ m/s [/tex] we have that

[tex]13.9 = \sqrt{ 2 g H}[/tex]

=> [tex]H = \frac{13.9^2}{2 * 9.8}[/tex]

=> [tex]H = 9.86 \ m [/tex]