If an object is thrown upward with an initial velocity of 38

If an object is thrown upward with an initial velocity of 384 ft/second, then its height after t seconds is given by the following equation. h = 384t - 32t^2 Find the maximum height attained by the object Find the number of seconds it takes the object to hit the ground The maximum height attained by the object is feet.

Solution

h = 384t - 32t^2

At max height velocity of object will be zero,

V = dh/dt = 0

dh/dt = 384 - 64*t = 0

t = 384/64 = 6 sec

Now at t = 6 height will be

h = 384*6 - 32*6^2 = 1152 ft

B.

when ball hits the ground h = 0,

So

0 = 384*t - 32*t^2

solving above equation

t = 0 OR t = 384/32 = 16 sec

So it will take 16 sec to object to hit the ground.