The height h, in feet, of a ball that is released 4 feet above the ground with an initial velocity of 80 feet per second is a fu
nction of the time t, in seconds, the ball is in the air and is given by
h(t)=-16t^2+80t+4, 0 < t < 5.04
a. Find the height of the ball above the ground 2 seconds after it is released.
b. Find the height of the ball above the ground 4 seconds after it is released.
Please show work!
2 answers:
Answer with Step-by-step explanation:
The height of the ball from the ground as a function of time is given by
The height of the ball at any instant of time can be found by putting the value of time 't' in the above relation as
Part a)
Height of ball after 2 seconds it is released is
Part b)
Height of ball after 4 seconds it is released is
Answer:
Part a)
Height of ball after 2 seconds it is released is
h(2)=-16\times 2^2+80\times 2+4=100feet
Part b)
Height of ball after 4 seconds it is released is
h(4)=-16\times 4^2+80\times 4+4=68feet
Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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ok so were starting with 7 and adding 3 every time
im going to do 3 times 19 like as if we started with 3 then im going to add 4
3 x 19 = 57
57 + 4 = 61
the 19th term is 61
Hi!!
1) 2/3 *12
You basically do this
2/3 * 12/1 = 24/3 which = 8!
2) 13 * 1/2
= 13/1 * 1/2
= 13/2
= 6 1/2
Hope this helps!
