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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?

User GRowing
by
6.6k points

1 Answer

7 votes

Answer:

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

User Mellet
by
7.0k points
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