The table shows the height of an object, h(t), in meters after t seconds. Use your calculator to find the quadratic function. Fi
nd the approximate time, to the nearest hundredth of a second, after which the object will reach its maximum height.
1 answer:
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Answer:
HG = 28
Step-by-step explanation:
The midsegment MN is half the sum of the 2 parallel bases, that is
= MN , that is
= 22 ( multiply both sides by 2 )
HG + 16 = 44 ( subtract 16 from both sides )
HG = 28
Answer:
(a) D(t)= -4.9 t^2 + vt
(b) 10 sec
(c) 122.5m
(d) At t= 4sec, v= 9.8m/s (upwards) and at t= 9sec, v= 39.2m/s (downwards)
(e) -24.5m
(f) -49m/s (i.e. downwards)
(g) 0
Step-by-step explanation:
(b) Time in the air = 2v/g = 10sec
(c) Maximum height = =122.5m
(d) Velocity at any time t = v -9.8t
So, at t= 4sec, velocity = 49 – 9.8(4) =9.8m/s
at t= 9sec, velocity = 49 – 9.8(9) = -39.2m/s
(e) Average velocity = = -24.5m
(g) the stone is thrown vertically upwards, so no horizontal distance covered.