Question

in the fu tbove the end -167 and and What is All play the action that the point the bal wher

Answer 1

Given the position of the object t seconds after the launch (which is also its height above the ground) is given by, s(t)= -16t? + v + So for t2 0. (a) From the given information, Initial velocity (v.) = 36 ft/s and so = 3 ft. Substitute above values in equation (1) to get the position of baseball. s(t)=-16° +36 +3 Therefore, the function that gives the position of baseball is s(t)=-16t? +36 +3

Graph of the function s(t)= -16° +36 +3 as shown below: Y 40+ $(t) = -16r2 + 367 +3 (b) Find the height of the ball when it reaches the high point. To get maximum height. Put = 0

(-16° + 36+ + 3) - (-168°) + (361) + . (3) =-32t +36 + 0 = -32t +36 Then, -32t +36 = 0 32t = 36 Substitute t = - sec in s(t)= -16° +36 +3. = 81 81 - — + — +3 4 2 -81+1623

+ 81+12 + = 23.25 ft Hence, the height of the ball when it reaches the high point is 23.25 ft.

(c) To find how long does the catcher have to wait to catch the ball. Put s(t)=0. -16t2 + 36t +3 = 0 -360 (36)2 - 4.(-16)(3) 2(-16) -36 V1296 +192 t= -32 _-36 + V1488 -32 -3614793 -32 3654193 32 Ol' - 36-4193, 36+4193 . 32 32 9-V93 9+ 193 - Ol 8 -0.08045 or 2.33045 Since t cannot be negative. Hence, at t = 2.33045 sec long does the catcher have to wait to catch the ball.