An object of mass 40 kg is raised to a height of 5 m above the ground. what isnits potential energy? if the abject is allowed to fall, find its kinetic energy when itbis half way down (g=10m/s)

Consider that I'm an italian student so I may unproperly use some terms.

Being Ug (Potential energy on earth) = m·g·h  

Where "m" in this case is the mass of the ball, "g" the average value of gravitation constant on Earth's surface and "h" the height reached by the ball.

The solution to the first question would be :  Ug = mgh = 40[kg] · 10 [m/s²] · 5[m] = 2000 [Joule]

In order to find the value of its K (kinetic energy) and without considering air resistance, you can use the concept of mechanical conservation of energy that confirms : U(start) + K(start) = U(final) + K(final).

So since you want to find its K(final) you can isolate it in the law and it gives you:

K(final) = U(start) + K(start) - U(final)

Consider that in halfway fall its Ug(final) value is mgh = 40[kg] · 10[m/s²] · 2.5[m] = 1000[Joule]

Now you just need to change these elements into their values(consider that K(start) is 0:

K(final) = 2000[Joule] + 0[Joule] - 1000[Joule] = 1000[J]

Hope I explained it well.

3.2x10^7

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer