A ball is thrown straight up with a initial velocity of 56 feet per second . The height , h , of the ball t seconds after it is
thrown is given by the formula h(t)=56t-16t^2 . What is the maximum height of the ball?
1 answer:
<span>Formula: H(t) = 56t – 16t^2
</span>
H(t) = - 16t^2 + 56t
<span><span>A.
</span>What is the height of the ball after 1 second? H
(1) = 56(1) – 16(1) ^2 = 40 pt.</span>
<span><span>B.
</span>What is the maximum height? X = - (56)/2(- 16) =
1.75 sec h (1.75) = 56(1.75) – 16(1.75) ^2 h (1.75) = 49ft.</span>
<span><span>C.
</span><span>After how many seconds will it return to the
ground? – 16t^2 + 56t = 0 - 8t =0 t = 0</span></span>
<span><span>-
</span><span>8t (2 + - 7) = 0 2t – 7 = 0 t = 7/2
Ans: 3.5 seconds</span></span>
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Good luck!!!
Answer:
$15.55
Step-by-step explanation:
12.50 x 2= 25
1.50 x 2= 3
18%=1.8 x 10(1.8)
25+3=28
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Answer= 15.55
Hope this helps.
