A rock is thrown vertically upward with a speed of 12.0m/s. Exactly 1.00s later, a ball is thrown up vertically along the same path with a speed of 18.0m/s. (a) At what time will they strike each other

Ok, remember that the only force acting on the objects will be the acceleration force.

First, let's describe the motion of the rock.

The acceleration will be:

Ar(t) = -9.8m/s^2

(where the negative sign is because the gravity pulls the rock downwards).

If we want to obtain the velocity of the rock, we should integrate over time and get:

Vr(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial velocity, and we know that it is:

v0  = 12.0m/s

Then:

Vr(t) = (-9.8m/s^)*t + 12m/s.

And the position can be obtained if we integrate again.

Pr(t) =  (-4.9m/s^)*t^2 + 12m/s*t + p0

Where p0 is the initial position.

Now let's do the same for the ball, but the ball is thrown one second after, so we will work with the variable (t - 1s)

Ab(t) = -9.8m/s^2

Vb(t) = (-9.8m/s^2)*(t - 1s) + 18m/s

Pb(t) =  (-4.9m/s^2)*(t - 1s)^2 + 18m/s*(t - 1s) + p0

As the ball is thrown along the same path than the rock, the initial position is the same in both cases.

Now the two objects will impact each other when:

Pb(t) = Pr(t)

(-4.9m/s^2)*(t - 1s)^2 + 18m/s*(t - 1s) + p0 =  (-4.9m/s^)*t^2 + 12m/s*t + p0

Let's solve this for t.

(-4.9m/s^2)*(t - 1s)^2 + 18m/s*(t - 1s) =  (-4.9m/s^)*t^2 + 12m/s*t

(-4.9m/s^2)*(t^2 -2*t*1s +  1s^2) + 18m/s*(t - 1s)  =  (-4.9m/s^)*t^2 + 12m/s*t

(-4.9m/s^2)*(-2*t*1s + 1s^2) + 18m/s*t - 18m = 12m/s*t

-4.9m + 9.8m/s*t + 18m/s*t  - 18m = 12m/s*t

-22.9m = t*(12m/s - 9.8m/s - 18m/s)

22.9m = t*16.8m/s

(22.9/16.8)s = 1.363 s

So the ball and the rock will strike each other at t = 1.363 seconds.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2