ww3.math msu.edu/webwork2/mth 132 8518 87800w13-2.7-Rates-of Change/7/key-bd26fa

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ww3.math msu.edu/webwork2/mth 132 8518 87800w13-2.7-Rates-of Change/7/key-bd26fa0425be236ae 325Sa1 Beua PreviousProblem List Next Help Entering Answers I Similar Example PDF (1 point) A ball is thrown upward from the top of a building 380 teet tall. The height of the ball is described by the function h(t) =-16t2 + 26t + 380, where t isthe time in seconds and t=0corresponds to the moment the ball is thrown (a) Determine for which value of t the ball reaches the maximum height and determine this maximum height Max Height ft e) Determiwebal rahes the ground (c) With what velocity does the ball hit the ground? Answer: Note: You can earn partial credit on this problem. Preview My Answers Submilt Answers Show me another You have attempted this problem 0 times You have 12 attempts remaining.

Explanation / Answer

h(t)=-16t2+26t+380

h'(t)=-32t +26

h'(t)=0

-32t +26=0

t=26/32=0.8125

So the ball reaches the maximum height at t=0.8125 sec

And maximum height, h(0.8125)=-16(0.8125)2+26(0.8125)+380=390.5625 feet

b. Ball reaches the ground means h(t)=0

-16t2+26t+380=0

t=5.7532 sec

c. velocity, h'(t)=-32t+26

at t=5.7532, velocity=-32(5.7532)+26=-158.1024 feet/sec

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