Question #af942

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Question #af942

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Answer 1 · 2017-03-13T16:44:41

$t = 3.6 s$ $t=3.6s$ , rounded to one decimal place.

Explanation:

Using FPS system of units.

The kinematic expression is
$h (t) = h_{0} + u t + \frac{1}{2} g t^{2}$ $h(t)=h_0+ut+1/2g t^2$
Inserting given values and noting that acceleration due to gravity $32 f t \cdot s^{- 2}$ $32ft cdot s^-2$ is acting against the direction of motion we get

$h (t) = 7 + 54 t + \frac{1}{2} (- 32) t^{2}$ $h(t)=7+54t+1/2(-32) t^2$
$h (t) = 7 + 54 t - 16 t^{2}$ $h(t)=7+54t-16 t^2$
Which is the given equation.
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.
We can start solution from here.

When the ball reaches ground $h = - 7 f t$ $h=-7ft$
Inserting in the given height equation we get
$- 7 = 7 + 54 t - 16 t^{2}$ $-7=7+54t-16 t^2$
$\Rightarrow 16 t^{2} - 54 t - 14 = 0$ $=>16 t^2-54t-14=0$
$\Rightarrow 8 t^{2} - 27 t - 7 = 0$ $=>8 t^2-27t-7=0$

Solving the quadratic with the inbuilt graphic tool we get two roots as below. Ignoring the $- v e$ $-ve$ root as time can not be negative we get

my comp
$t = 3.6 s$ $t=3.6s$ , rounded to one decimal place.

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Question #af942

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