Aprojectile is thrown upward so that its distance above the ground after t seconds is h (t )equals negative 16 t squared plus 704 t. a) find the maximum height the object reaches. b) find the time it takes the object to reach the ground. (hint: when an object reaches the ground, its height is 0.)

a) The maximum height the projectile reaches is 7744 m.

b) The time it takes the object to reach the ground is 44 s.

Explanation:

Hi there!

The height function is the following:

h(t) = -16 t² + 704 t

a) Let´s find the time it takes the projectile to reach the maximum height. For this, we can use the fact that at the maximum height, the velocity of the projectile is zero. The velocity is the variation of height with respect to time, in other words, it is the derivative of the height function:

v = dh/dt = -2 · 16 t + 704 = -32 t + 704

At the maximum height:

v = 0

0 = -32 t + 704

-704/-32 = t

t = 22 s

Now, we can calculate the height at time t = 22 s:

h(t) = -16 t² + 704 t

h(22) = -16 (22)² + 704 (22)

h(22) = 7744 m

The maximum height the projectile reaches is 7744 m.

b) When the object reaches the ground h(t) = 0. Then:

h(t) = -16 t² + 704 t

When the object reaches the ground:

0 = -16 t² + 704 t

Let´s solve this quadraic equation for "t":

0 = -16 t² + 704 t

0 = t (-16 t + 704)

t = 0 s

and

-16 t + 704 = 0

-16 t = -704

t = -704/-16

t = 44 s

The time it takes the object to reach the ground is 44 s.

newton's first law of motion

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