If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t2. (a) What is the maximum height reached by the ball? ft (b) What is the velocity of the ball when it is 384 ft above the ground on its way up? (Consider up to be the positive direction.) ft/s What is the velocity of the ball when it is 384 ft above the ground on its way down?

a) Smax = 400 ft

b) v = 32 ft/s

c) v = - 32 ft/s

Explanation:

(a)

The function given for the height of ball is:

s = 160 t - 16 t²

Therefore, in order to find the time to reach maximum height (in-flexion point), we must take the derivative with respect to t and set it equal to zero:

Therefore,

160 - 32 t = 0

32 t = 160

t = 160/32

t  = 5 sec

Therefore, maximum distance is covered at a time interval of 5 sec.

Smax = (160)(5) - (16)(5)²

Smax = 800 - 400

Smax = 400 ft

b)

First we calculate the time at which ball covers 384 ft

384 = 160 t - 16 t²

16 t² - 160 t + 384 = 0

Solving Quadratic Equation:

Either:

t = 6 sec

Or:

t = 4 sec

Since, the time to reach maximum height is 5 sec. Therefore, t < 5 sec

Therefore,

t = 4 sec

Now, we find velocity at 4 sec by taking derivative od s with respect to t at 4 sec:

v = 160 - 32 t

v = 160 - (32)(4)

v = 32 ft/s

c)

Since, t = 6 s > 5 s

The second value of t = 6 sec must correspond to the instant when the ball is 384 ft above the ground while traveling downward.

Hence, velocity at that time will be:

v = 160 - (32)(6)

v = -32 ft/s

Negative sign due to downward motion.

hey , the answer is c

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