SOLUTION: Bob throws a ball vertically upward from the top of the cliff, the height of the ball above the base of the cliff is approximated by the model h = 6s + 10t- 5t ^2 where h is the he
Question 694546: Bob throws a ball vertically upward from the top of the cliff, the height of the ball above the base of the cliff is approximated by the model h = 6s + 10t- 5t ^2 where h is the height in metres and t is the time in seconds.
a) How high is the cliff?
b) How long does it take the ball to reach a height of 50 metres above the cliff?
c) After how many seconds does the ball hit the ground?
can you please help me out? I really don't understand. Thanks in advance :) Answer by lynnlo(4176) (Show Source):