An object is thrown straight up with a velocity, in ft/s, given by v(t)=−32t+79v(t)=−32t+79, where tt is in seconds, from a height of 48 feet. a) What is the object's initial velocity? b) What is the object's maximum velocity? c) What is the object's maximum displacement? d) When does the maximum displacement occur? e) When is the object's displacement 0? f) What is the object's maximum heig

a) , b) , c) , d) , e) , f)

Step-by-step explanation:

a) Initial velocity

b) Maximum velocity

Due to the negative slope and positive intercept, it is concluded that object's maximum velocity is the initial velocity found in previous point.

c) Maximum displacement

First, the displacement function, measured in feets, has to be found by integrating the velocity function:

The initial condition is . Then:

The complete displacement function is presented as follows:

Besides, the acceleration function is determined by deriving the velocity function:

This finding allows to infer the existence of a maximum displacement by applying Second Derivative Test. The critical point is found by equalizing the velocity function to zero and clearing time:

Finally, the maximum displacement is obtained:

d) The maximum displacement occur at .

e) Let equalize to zero and the two roots are found by using Second-Order Polynomial General Equation:

Where and . is the only solution that is physically reasonable.

f) Maximum height

The maximum height is given by the maximum displacement obtained in point d). Hence,

its 4 because 2 times 2 is 4, if you double the number two you get four

the answer is mp

step-by-step explanation: