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Sagot :
Answer: hi there is think that I can help you with your problem!
h = -16t^ 2 + 20t + 6 ....the ball will hit the ground when h = 0
0 = -16t^2 + 20t + 6 factor
0 = ( -8t - 2 ) ( 2t - 3) set each factor to 0
-8t - 2 = 0 and 2t - 3 = 0
-8t = 2 2t = 3
t = -2/8 = - 1/4 sec [reject] t = 3/2 sec = 1.5 sec
Max ht will be reached at [ -b / 2a ] sec = -20 / [ 2(-16)] = 20/32 = 5/8 sec
Step-by-step explanation:
-16t^2+20t+6 = 0
will give you the times that the ball is on the ground. The negative root accounts for the fact that the ball is 6 feet above ground at t=0.
I get t=3/2 and t=-1/4
-2(2t -3) (4t +1) = 0
2t -3 = 0
2t-3 + 3 = 0 + 3
2t = 3
t = 3/2 second
I hope that I helped you and have a nice day.
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