An object is thrown vertically upward from a height of $2 m$ $2 m$ at $1 \frac{m}{s}$ $1 m/s$ . How long will it take for the object to hit the ground?

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An object is thrown vertically upward from a height of $2 m$ $2 m$ at $1 \frac{m}{s}$ $1 m/s$ . How long will it take for the object to hit the ground?

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Answer 1 · 2017-01-01T02:09:00

The initial velocity of the object in upward direction $u = 1 \frac{m}{s}$ $u=1m/s$
( upward direction is taken positive )

So the downward acceleration due to gravity $g = - 9.8 \frac{m}{s^{2}}$ $g=-9.8m/s^2$

The object is thrown from height $2 m$ $2m$ . If it reaches ground after $t s$ $ts$ then the displacement of the object during $t s$ $t s$ will be $h = - 2 m$ $h=-2m$
Hence applying equation of kinematics for under gravity motion we can write
$h = u \times t + \frac{1}{2} \times g \times t^{2}$ $h=uxxt+1/2xxgxxt^2$

$\Rightarrow - 2 = 1 \times t - \frac{1}{2} \times 9.8 \times t^{2}$ $=>-2=1xxt-1/2xx9.8xxt^2$

$\Rightarrow 4.9 t^{2} - t - 2 = 0$ $=>4.9t^2-t-2=0$

$\Rightarrow t = \frac{1 + \sqrt{{(- 1)}^{2} - 4 \times 4.9 \times (- 2)}}{2 \times 4.9} \approx 0.75 s$ $=>t=(1+sqrt((-1)^2-4xx4.9xx(-2)))/(2xx4.9)~~0.75s$

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An object is thrown vertically upward from a height of $2 m$ $2 m$ at $1 \frac{m}{s}$ $1 m/s$ . How long will it take for the object to hit the ground?

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An object is thrown vertically upward from a height of $2 m$ $2 m$ at $1 \frac{m}{s}$ $1 m/s$ . How long will it take for the object to hit the ground?