Question

A 5kg bag falls a verticle height of 10m before hitting the ground.

Ek=1/2mv^2 and Ep=m g h (assume g=9.8N/kg).
Using these two equations, calculate the speed the bag reaches before hitting the ground.

Answer 1

`u = 7m {s}^( - 1)`

Step-by-step explanation:

We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:

`me(1) = me(2)`

where me(1) is mechanical energy while on h=10m

and me(2) is mechanical energy while on the ground

Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)

Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.

DynamicE(2) is equal to zero since it's touching the ground

Using that info we have

`m * g * h = (1)/(2) * m * u {}^(2) \\`

we divide both sides of the equation with mass to make the math easier.

`9.8 * 10 = (1)/(2) * u {}^(2) \\ (98)/(2) = u {}^(2) \\ u { }^(2) = 49 \\ u = 7`