On May 20, 1999 Robbie Knievel easily cleared a narrow part...

Name: On May 20, 1999 Robbie Knievel easily cleared a narrow part of the Grand Canyon during a world-record-setting long-distance motorcycle jump: a horizontal distance of 69.5 m. He left the jump ramp at a 10° angle above the horizontal. How fast was he traveling when he left the ramp? Indicate any assumptions you made.
Uploaded: 2020-10-17T16:26:17-08:00
Duration: 5 min 7 s
Channel: Manish Kumar
Description: On May 20, 1999 Robbie Knievel easily cleared a narrow part of the Grand Canyon during a world-record-setting long-distance motorcycle jump: a horizontal distance of 69.5 m. He left the jump ramp at a 10° angle above the horizontal. How fast was he traveling when he left the ramp? Indicate any assumptions you made.

On May 20, 1999 Robbie Knievel easily cleared a narrow part of the Grand Canyon during a world-record-setting long-distance motorcycle jump: a horizontal distance of 69.5 m. He left the jump ramp at a 10° angle above the horizontal. How fast was he traveling when he left the ramp? Indicate any assumptions you made.

Transcript

00:01 So we will assume that robbie begins his jump at the same elevation as he learned if we know motion into two phase tnor to t1 and then t1 to t2.

00:16 We are assuming our motion into two phase and actually motion into two phase t1 to t1 and where t1 is the time when roby at the maximum attitude and t1 to t2 up to the end of the motion.

00:30 Ignoring the effect of the air resistances our no no so basically in the non -information over there is x nor v0 obviously and t nor over there is zero y nor initial height also and that means and also vnore x equals to v1 x is equals to v2 x because there is no x -lidation in the direction so and this is equals to v node course of 10 degree meter per second and anode x is obviously 0 or a x directly is 0 meter per second v nor y is obviously v nor sign of 10 degree and v1y is 0 meter per second at the maximum altitude obviously and y2 over there that will be 0 as well and over there x2 which is 69 .5 meter and and unknown over there is v nor obviously and ay we already know that the acceleration in the y direction is what minus of 9 .8 obviously in the downward direction.

01:42 So we have to calculate vnor over there.

01:44 So by the given information applying the kinematics law for the equation in the xy, so x2 over there that will be nothing but x nor plus vnor x into t2 and that is even which is equals to 69 .5 so and over there x nor is zero so zero plus v nor and x which is v nor cost of 10 degree okay see over there this thing over there is zero so basically we just going to put over there so v node x is just b nor cost 10 degree into t2 which is equals to 69 .5 and in the direction of y in the direction of y in the direction of y be 1y would be nothing but you v nor y you plus a t a nor y into t which is also equals to 0 and that if we're gonna just put all the values of these things over there is nothing what v nor sign of 10 degree the sign of 20 degree yeah and plus of an nor is nothing but minus of 9 .8 into t 1 actually actually over the 1 is equal to 0 so yeah so the vertical position of the font size is systematic parabola in the parameter time from the which t2 to t2 to t2 the w, which is maximum altitude in the half of the time interval and it takes complete jump.

03:18 So you can say by the ehobe equation that vnor sign of 10 degree is just equals to minus of 9 .8 upon t2 by 2.

03:33 Why is that because it takes the same time to you know goes to the maximum height and land to the ground level or so that's the same time so t t1 is just the two times of t2 so that means t1 is just t 2 upon t2 upon 2 so that's is equal to zero substituting equations over there so v nor sign 10 degree which is minus of 9 .8 upon 2 and we have vibrated expression for t2 over there so t 2 would be nothing but 69 .5 upon 39 .5 upon v...

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