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A motorcycle stunt rider rides off …

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A motorcycle stunt rider rides off the edge of cliff. Just at the edge his velocity is horizontal with magnitude 9m/s . Find the magnitude of the motorcycles position vector after 0.50s it leaves the edge of the cliff

Posted by Shrishti Mishra 3 years ago

CBSE > Class 11 > Physics

1 answers

Ankit Kumar Sain 3 years ago

Solution verified Verified by Toppr At t=0..50s the x−and y- coordinates are x=v 0x t =(9.0ms −1 )(0.50s)=4.5m y=− 2 1 gt 2 =− 2 1 (10ms −2 )(0.50s) 2 =− 4 5 m the negative value of y shows that this time the motorcycles is below its starting point. The motorcycle's distance from the origin at this time. r= x 2 +y 2 = ( 2 9 ) 2 +( 4 5 ) 2 = 4 349 m The components of velocity at this time are v x =v 0x =9.0ms −1 v y =−gt=(−10ms −2 )(0.50s)=−5ms −1 The speed (magnitude of the velocity ) at this time is v= v x 2 +v y 2 = (9.0ms −1 ) 2 +(−5ms −1) 2 = 106 ms −1

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