Question

A rock is thrown upward with an initial velocity of 122 feet per second. The height of the rock ,(h), in feet after t seconds is given by the function ℎ() = −162 + 122 + 10.How long does it take the rock to reach its maximum height? What is the rock’s maximum height. Round to the nearest hundredth, if necessary.

Answer 1

i have no clue with this one help −4(3) + 2(−); for = 1/2, = −3

Answer 2

Answer: the maximum height is 231.118 feet

Explanation:

Initial velocity, u = 122 feet per second.

The height of the rock ,(h), in feet after t seconds is given by the function ℎ(t) = −162x^2 + 122x + 10.

Acceleration of the object is 32.17405 ft/s2. This is approximately 32.2 ft/s2. It is same as the acceleration due to gravity. Since the object is moving upwards, the acceleration will be negative because it is moving in the opposite direction of the gravitational force.

Therefore

g = - 32.2 ft/s^2

The final velocity, v = 0 at maximum height.

Applying Newton's equation of motion,

v^2 = u^2 + 2gh

0 = 122^2 + (-32.2 × 2 ×h)

0 = 14884 - 64.4h

64.4h = 14884

h = 14884/64.4 = 231.118 feets( approximated to the nearest hundredth)