Aball is thrown from the top of a building. at time t (in seconds), its height (in feet) follows the path s(t) = -3t2 + 24t + 60.
a. when does it hit the ground?
b. what was its acceleration at time t = 1 ?

10 sec

Explanation:

Given

Ball follows the path

Therefore at t=0

S(0)=60

i.e. height of building is 60 m

(a)Ball will hit when the distance between ball and ground is zero i.e.

i.e. at t=10 sec ball will hit ground

(b)accleration at time t=1 sec

acceleration is given by

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.