Question #9a724

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Answer 1 · 2018-02-06T22:40:56

$3.5 s$ $3.5s$

graph{-16x^2+54x+7 [-5, 5, -5, 5]}

when the ball hits the ground, its height is $0$ $0$ .

$h (t) = 0$ $h(t) = 0$

this means that $- 16 t^{2} + 54 t + 7 = 0$ $-16t^2 + 54t + 7 = 0$ .

$(8 t + 1) (2 t - 7) = 16 t - 56 t + 2 t - 7 = 16 t - 54 t - 7$ $(8t+1)(2t-7) = 16t - 56t + 2t - 7 = 16t - 54t - 7$

$- (8 t + 1) (2 t - 7) = - 16 t + 54 t + 7$ $-(8t+1)(2t-7) = -16t + 54t + 7$

factorising gives $- (8 t + 1) (2 t - 7) = 0$ $-(8t+1)(2t-7) = 0$ .

$0 = - 0$ $0 = -0$

$(8 t + 1) (2 t - 7) = 0$ $(8t+1)(2t-7)=0$ .

$8 t = - 1$ $8t = -1$ or $2 t = 7$ $2t = 7$

$t = - \frac{1}{8}$ $t = -1/8$ or $t = \frac{7}{2}$ $t = 7/2$

note that the answer asks for the height after $t$ $t$ seconds. this means that only the positive solution of $t$ $t$ applies.

here, this is $\frac{7}{2}$ $7/2$

$\frac{7}{2} = 3.5$ $7/2 = 3.5$

$3.5$ $3.5$ seconds after being thrown up from $7$ $7$ feet above ground, the ball will reach the ground again (assuming that the ground is level).