Question

5. A ball is thrown upward with an initial velocity of 50 feet per second. The

height h (in feet) of the ball after t seconds is given by h =
= -16t2 + 50t. At
the same time, a balloon is rising at a constant rate of 20 feet per second. Its
height h in feet after seconds is given by h = 20t.

a. When do the ball and the balloon reach the same height?
b. When does the ball reach its maximum height?
c. When does the ball hit the ground?

Answer 1

SEE BELOW IN BOLD.

Explanation:

a.

h = -16t^2 + 50t

h = 20 t

When the height is the same:

-16t^2 + 50t = 20t

-16t^2 + 30t = 0

t(-16t + 30) = 0

t = 0 or -16t + 30 = 0, so:

t = 0 or -30/-16 = 1.875

So the answer is 1.88 seconds to the nearest hundredth.

b.

For the ball

h = -16t^2 + 50t

Finding the derivative and equating to zero:

dh/dt = -32t + 50 = 0

t = -50/-32 = 1.563

Maximum height after 1.56 seconds to nearest hundredth

c.

When the ball hits the ground h = 0 so

-16t^2 + 50t = 0

-16t(t - 50/16)= 0

T = 3.13 SECONDS TO THE NEAREST HUNDERDTH