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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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A soccer ball is kicked horizontally off a 22.0 meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

$\begin{array}{1 1}16.5m/s\\17.5m/s\\18.5m/s\\19.5m/s\end{array} $

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Answer : 16.5m/s
Given:
$x=35.0m$
$a_x=0m/s/s$
$y=-22.0m$
$v_{iy}=0m/s$
$a_y=-9.8m/s/s$
Use $y=v_{iy}.t+0.5\times a_y\times t^2$ to solve for time $\rightarrow $ the time of flight is 2.12 seconds.
Now use $x=v_{ix}\times t+0.\times a_x\times t^2$ to solve for $v_{ix}$.
Note that $a_x$ is 0m/s/s so the last term on the right side of the equation cancels.
By substituting 35.0m for x and 2.12s for t the $v_{ix}$ can be found to be 16.5m/s
answered Aug 13, 2014 by sreemathi.v
edited Aug 13, 2014 by balaji.thirumalai
 

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