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3 years ago

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If an object is thrown upward at 128 feet per second from a height of 76 feet, its height S after t seconds is given by the foll

owing equation. S(t) = 76 + 128t − 16t2 (a) What is the average velocity (in ft/sec) in the first 4 seconds after it is thrown? ft/sec (b) What is the average velocity (in ft/sec) in the next 4 seconds?

1 answer:

faltersainse [42]3 years ago

3 0

<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>

Explanation:

a) Given S(t) = 76 + 128t − 16t²

s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft

s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

Displacement in 4 seconds = 332 - 76 = 256 ft

Time = 4 - 0 = 4 s

$\texttt{Velocity = }\frac{256}{4}=64ft/s$

Average velocity in first 4 seconds is 64 ft/s upward

a) Given S(t) = 76 + 128t − 16t²

s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft

Displacement in 4 seconds = 78 - 332 = -254 ft

Time = 4 - 0 = 4 s

$\texttt{Velocity = }\frac{-254}{4}=-63.5ft/s$

Average velocity in second 4 seconds is 63.5 ft/s downward

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<h3>What is an escape velocity?</h3>

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Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

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