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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Motion in a Plane

In a particular laboratory experiment,a spring gun placed on a table fires a steal ball horizontally outward.A student determines that the ball starts 1.0m above the floor and travels 2.7m horizontally before it strikes the floor.Determine the time that the ball is in the air and initial velocity of the ball

$\begin{array}{1 1}t=0.45s,\quad v_0=6.0m/s\\t=0.90s,\quad v_0=6.0m/s\\t=0.90s,\quad v_0=3.0m/s\\t=0.45s,\quad v_0=3.0m/s\end{array} $

1 Answer

Answer : $t=0.45s,\quad v_0=6.0m/s$
x=2.7m
y=-1.0m
$v_{0x}=v_0\cos \theta=v_0$
$v_{0y}=v_0\sin \theta=0$
$y=v_{0y}t+\large\frac{1}{2}$$(-g)t^2$
$\Rightarrow t=0.45s$
$x=v_{0x}t$
$\Rightarrow v_{0x}=v_0=6.0m/s$
answered Aug 13, 2014 by sreemathi.v
 

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