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

A player kicks a football at an angle of $30^{\circ}$ with horizontal and initial velocity of 19.6m/s. A second player standing at a distance of 20.95m from the first player and in the direction of the kick,starts running to meet the ball, at the instant it is kicked. How far and how fast must he run in order to catch the ball before it hits the ground ?

\[(a)\;12 m,6 m/s\quad (b)\;14m,6.5m/s \quad (c)\;13m , 8 m/s \quad (d)13m,6.5 m/s\]

1 Answer

Let the ball be kicked from P and seconds player be at Q
Ball and second player reach O at same time,t,
Therefore $t=\large\frac{2u \sin \theta}{g}$
$\qquad=\large\frac{2 \times 19.6 \times \sin 30}{9.8}$
$\qquad=2s$
$PO=R=\large\frac{u^2 \sin 2 \theta}{g}$
$\qquad=\large\frac{(19.6)^2 \sin 60}{9.8}$$=33.95m$
Distance covered by $2 ^{nd}$ player is
$33.95-20.95=13m$
Speed of the second player
$\qquad=\large\frac{13}{2}$
$\qquad=6.5m/s$
Hence d is the correct answer.

 

answered Jun 28, 2013 by meena.p
edited Jul 24, 2014 by thagee.vedartham
 

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