A ball is thrown down vertically with an initial speed of v0 from a height of h. (a) What is its speed just before it strikes the ground? (b) How long does the ball take to reach the ground? What would be the answers to (c) part a and (d) part b if the ball were thrown upward from the same height and with the same initial speed? Before solving any equations, decide whether the answers to (c) and (d) should be greater than, less than, or the same as in (a) and (b).

a)   v² = v₀² + 2 g h,  b)   t = v₀/g  (1+ √ (1 + 2gh/ v₀²))

Explanation:

a) This is an exercise that we can solve using conservation of energy.

Starting point. High point

         Em₀ = K + U = ½ m v₀² + m gh

Final point. Soil

          = K = ½ m v²

energy is conserved because there is no friction

         Emo = Em_{f}

         ½ m v₀² + m g h = ½ m v²

         v² = v₀² + 2 g h

b) the time it takes to reach the ground can be calculated with kinematics

let's create a reference frame with positive upward direction

         v = vo - g t

when it reaches the ground it has a velocity v, the initial velocity is downwards v₀ = -v₀

        v = -v₀ - gt

        t = - (v + v₀) / g

we substitute the velocity values calculated in the previous part

        t = - (√(v₀² + 2 g h) + vo) / g

we will simplify the equation a bit

        t = - v₀/g  (1+ √ (1 + 2gh/ v₀²))

c) is now thrown vertically upward with the same initial velocity vo.

   To find the final velocity we use the conservation of energy where the velocity is squared, so it does not matter if it is positive or negative, therefore in this section the value should be the same as in part a

         v = √ (v₀² + 2gh)

d) for this part if there is change since the speed is not squared

     v₀ = v₀

          v = v₀ - gt

          t = (v₀ - v) / g

          t = (v₀ - √(v₀² + 2 g h)) / g

          t = v₀/g   (1 - √(1 + 2gh / v₀²))

the answer is false

explanation:

increase masses, decrease separation