How many seconds will the ball be going upward if a ball is thrown vertically upward from the ground with an initial velocity of 20 feet per second and $a (t) = - 32$ $a(t)= -32$ feet per second squared?

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How many seconds will the ball be going upward if a ball is thrown vertically upward from the ground with an initial velocity of 20 feet per second and $a (t) = - 32$ $a(t)= -32$ feet per second squared?

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Answer 1 · 2015-03-13T16:42:47

The ball will continue to go upward until its (upward) velocity $= 0$ $= 0$

The ball's upward velocity is the sum of
its initial upward velocity
plus its upward velocity (really minus its downward velocity) due to gravity.

This velocity will $= 0$ $=0$ for a time $t$ $t$ (in seconds) when
$0 = \frac{20 f t}{sec} + \frac{- 32 f t}{{sec}^{2}} \cdot t sec$ $0 = (20 ft)/(sec) + (-32 ft)/(sec^2)* t sec$

This simplifies to
$t = \frac{5}{8} sec .$ $t = 5/8 sec.$

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How many seconds will the ball be going upward if a ball is thrown vertically upward from the ground with an initial velocity of 20 feet per second and $a (t) = - 32$ $a(t)= -32$ feet per second squared?

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How many seconds will the ball be going upward if a ball is thrown vertically upward from the ground with an initial velocity of 20 feet per second and a(t)=−32a(t)= -32 feet per second squared?

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How many seconds will the ball be going upward if a ball is thrown vertically upward from the ground with an initial velocity of 20 feet per second and $a (t) = - 32$ $a(t)= -32$ feet per second squared?