### Dr. Wilson

3. Given

Find formulas for v(t) and y(t). Include your constants of integration. Interpret these constants of integration.

Using these formulas, find when a water balloon which is launched upward with an initial velocity of 50ft/sec from the top of a stadium, which is 40ft above the ground, hits the ground. What is its velocity when it hits the ground?

To find the velocity as a function of time, start with the differential equation

dv/dt = -32

(we can suppress the units. They will automatically come out right if we proceed correctly.)

Separate the variables

dv = -32dt

Integrate both sides

v(t) = -32t + C

We can determine the constant of integration if we let t = 0. Our equation then becomes

v(0) = -32(0) + C

and

C = v(0)

The velocity when t = 0. If we let v(0) = vo, then our equation for the velocity becomes

v(t) = -32t + vo

where , the velocity when t = 0 is a constant.

To find the position, we use the differential equaton

ds/dt = -32t + vo

Separate the variables

ds = (-32t + vo)dt

Integrate both sides

s = -32t2/2 + vot + K

= -16t2 + vot + K

where K is a second constant of integration. To find out what K is, let t = 0 again.

s(0) = -16(0)2 + vo(0) + K = K

So K = s(0), the position when t = 0. If we let s(0) = so, then the position becomes

s = -16t2 + vot + so

To find when a water balloon which is launched upward with an initial velocity of 50ft/sec from the top of a stadium, which is 40ft above the ground, hits the ground, we let

vo = 50 and so = 40

and get

s = -16t2 + 50t + 40

To find the time when the water balloon hits the ground, we use the fact that s = 0 on the ground and get

-16t2 + 50t + 40 = 0

Since this is a quadratic, we can use the quadratic formula and get

The value of t that we get with the negative square root is approximately -0.66, a negative number. This reflects the fact that If the water balloon had been launched from below the stadium so that it reached the top of the stadium at time t = 0 with a velocity at that point of 50 ft/sec. upward, then it would have been at the level of the ground approximately 0.66 sec. before we started measuring time when the balloon left the top of the stadium. However, this model does not apply to times before the water balloon was launched, so we conclude that it hits the ground after approximately 3.76 sec.

To find its velocity when it hits the ground we plug this value for t into the formula for the velocity as a function of t.

The negative velocity indicates that the water balloon is heading down when it hits the pavement on the ground.