Question

To catch a ball, a professional baseball player leaps into the air with an initial velocity of

Answer 1

Answer:

The answer is below

Step-by-step explanation:

a) The baseball player leaps into the air with an initial velocity of  14 feet per second. This means that the player leaps from the ground, the initial height is therefore zero.

The height (y) is given by the formula:

y(t) = ut - (1/2)gt² + initial heigth

u = initial velocity = 14 ft/s, t = time taken, g = acceleration due to gravity = 32 ft/s². Substituting:

y(t) = 14t - (1/2) * 32 *t² + 0

y(t) = 14t - 16t²

b) when the player is on the ground, the height = 0. hence:

0 = 14t - 16t²

16t² - 14t = 0

t(16t - 14) = 0

t = 0 or 16t - 14 = 0

t = 0 or 16t = 14

t =0 or t = 0.875

Hence t = 0.875 seconds

c) The domain is the set of possible values for the time. Hence:

Domain = (0, 0.875] = 0 < t ≤ 0.875