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a ski jumper starts with the horizontal take off velocity of 25m/s and lands on a straight landing hill inclined at 30 o . Determine (a) the time between take-off and landing, (b) the length d of the jump, © the maximum vertical distance between the jumper and the landing hill. a ski jumper starts with the horizontal take off velocity of 25m/s and lands on a straight landing hill inclined at 30o. Determine (a) the time between take-off and landing, (b) the length d of the jump, © the maximum vertical distance between the jumper and the landing hill.
While drawing the free body diagram of the given problem , you will notice that at the moment the jumper is in the air....at that instant, effective acceleration due to gravity would be g*sin600 and along the incline there would be a component of g ,i.e, g*cos600 . After resolving the g into components let us now divide the velocity as such. We will see that along the incline the component of velocity is v*cos300 whereas acting opposite to the effective acceleration due to gravity is a component of v ,i.e., v*sin300 . Successfully resolving these into components we can now calculate the time bettween takeoff and landing as Initial velocity in upwards direction being v*sin300 that is, 12.5m/s. Effective g is g*sin600. So, by veffective=ueffective-geffective*t where veffective=0m/s ; ueffective= 12.5m/s; geffective=g*sin600we get the required time as t=1.47*2=2.94 secondsLength d of the jump can be calculated by s=ueffective*t+0.5*acceleration*t2 here u effective =v*cos300=21.65m/s;t=2.94 s;acceleration=g*cos60=4.9m/s2so s=d=1948.94m along the inclined plane....if d is measured with respect to ground then that will be dcos300 =1687.78mmaximum distance between the jumper and hill would be given by veffective2 =ueffectrive2-2*geffective*h where veffective=0m/s;ueffective=12.5m/s;geffective=8.49m/s2 so h= 9.20 mI guess there are no minor mistakes,,, if there were the free body diagram should be clear enough and you may do your calculations yourself. Thank you...
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