Question 4 (1 point) A ball is thrown from a mountain 960 feet high, into the air with an upward velocity of 64 ft/s. When did the ball hit the

ground? And What was the max height it reached?

The ball will hit the ground at 10 seconds and its maximum height is 1024 feet.

Step-by-step explanation:

Height of a mountain, h = 90 feet

Upward velocity of a ball, u = 64 ft/s

Its height as a function of time can be written by using the second equation of motion as follows :

.....(1)

When it will hit the ground, f(x) = 0

It is quadratic equation.

t = -6 s and t = 10 s

Neglecting -6 seconds, the ball will hit the ground at 10 seconds.

For maximum height,

Put f'(x) = 0

Put t = 2 in equation (1)

Hence, the ball will hit the ground at 10 seconds and its maximum height is 1024 feet.

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

they can both be solutions depends on the equation

step-by-step explanation:

what is the equation? ?