Question

an object is thrown upwards from a height of 40.0 m. A second object is dropped from that height 3.0 s later. What must be the initial speed of the first ball if both are to hit the ground at the same time?

Answer 1

Let y1(t) and y2(t) be the heights of the first and thesecond balls at moment t respectively.
Let also v be the initial velocity of the first ball, andh1=40m be the initial height.
Then v1 is directed up.
Since both ball move with constant accelerationg=9.8m/s^2 directed down, we have that

y1(t) = h1 + v*t- g*t^2/2


y2(t) = h1, for t < 3s
= h1 -g*(t-3)^2 / 2, for t > 3s

Let s be the moment at which both balls hit the ground.
Then
y1(T) = y2(T) =0.

Hence from
y2(T) = 0
we get
h1 - g*(T-3)^2/ 2 = 0
g*(T-3)^2 / 2 =h1

T = 3 +square_root(2*h1/g)
= 3 +square_root(2*40/9.8)
= 5.8571seconds


Substitute the value of s into the identity:
y1(T) = 0

h1 + v*T -g*T^2/2 = 0

v*T = g*T^2/2 -h1

v = g*T/2 -h1/T
=9.8*5.8571/2 - 40/5.8571
= 21.870 m/s

Answer: v = 21.870 m/s