Question

A baseball throwing machine throws a baseball straight up with an initial velocity of 32 ft/sec from a height of 4.5 ft.

(a) Find an equation that models the height, h, of the ball t seconds after it is thrown.

A) h(t) = -16t^2 + 32t + 4.5
B) h(t) = 16t^2 + 32t - 4.5
C) h(t) = 16t^2 - 32t + 4.5
D) h(t) = -16t^2 - 32t - 4.5

Answer 1

The correct equation modeling the height h(t) of the baseball t seconds after it has been thrown straight up with an initial velocity of 32 ft/sec from a height of 4.5 ft is A) h(t) = -16t^2 + 32t + 4.5.

Step-by-step explanation:

The problem involves finding the height h(t) of a baseball at any time t seconds after it has been thrown straight up with an initial velocity. When the only force acting on the baseball is gravity (ignoring air resistance), the motion of the baseball can be modeled by the equation of motion:

h(t) = -16t^2 + vt + s,

where h(t) is the height at time t, v is the initial velocity, and s is the initial height. For this problem, the initial velocity v is 32 ft/sec, and the initial height s is 4.5 ft, resulting in the equation:

h(t) = -16t^2 + 32t + 4.5.

The correct choice is A) h(t) = -16t^2 + 32t + 4.5. This equation fits the model for an object thrown upward, where the acceleration due to gravity is a downward force represented by -16 ft/s2 in the US customary system of measurement.