How many seconds will the ball be going upward if a ball is thrown vertically upward from the ground with the initial velocity of 56 feet per second and the acceleration due to gravity is $- 32 \frac{f t}{t^{2}}$ $-32 (ft)/t^2$ ?

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Answer 1 · 2015-03-01T19:17:59

The ball will be going up for 1.75s

$v = u + a t$ $v=u+at$

$v$ $v$ = final velocity which is zero in this case

$u$ $u$ =initial velocity

$a$ $a$ =acceleration = 32ft/s/s

$t$ $t$ = time

$0 = u + a t$ $0=u+at$

$t = \frac{- u}{a} = \frac{- 56}{- 32} = 1.75 s$ $t=(-u)/(a) =(-56)/(-32)=1.75s$

How many seconds will the ball be going upward if a ball is thrown vertically upward from the ground with the initial velocity of 56 feet per second and the acceleration due to gravity is −32ftt2-32 (ft)/t^2?