Astone is thrown vertically into the air from a tower 110 ft. high at the same time that a second stone is thrown upward from the ground. the initial velocity of the first stone is 60 ft/s and that of the second stone 85 ft/s. when and where will the stones be at the same height from the ground?

The stones will be at the same height of 62.59 feets 4.4 seconds later.

Explanation:

We will be using the following kinematic equation for 1D movement:

.

For the first stone we got:

,

of course, taking the gravitational acceleration

.

For the second stone we got:

.

.

Now, we want to find the time t' at which both stones will be at the same height, this is

.

So

.

Working a little the equation

Now, to find where the stones will be, we can just replace t for t' in any of the two formulas for the stones:

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//