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        A ball is projected from a certain point on the surface of planet at an angle with horizontal surface. the horizontal and vertical displacement x and y vary with time in second as : x=10√3t and y=10t-t^2.the maximum height attained isA.100m.      B.75m.     C.50m.     D.25m
2 years ago

							D.25m is the correct answer.Using vector equation ​ where ​ is acc. due to grav. on the planet, we get :  ​ = ​ . On comparing and evaluating corresponding terms, we get    so,  ​ , |u|=20m/s and  .From ​, max. height=25 m.

2 years ago
							From  x=10√3t and y=10t-t^2substitute the t value in y eqn. we get,y = x/root(3) – x^2/300on comparing with the trajectory eqn. of motion we get thetha = 30 degree and on solving for we get u = 20root(5) m/sMaimum height = u^2sin^2thetha/g => 2000/4g => 50 meter.

2 years ago
							From  x=10√3t and y=10t-t²substitute the t value in y eqn. we get,y = x/√3– x²/300on comparing with the trajectory eqn. of motion we get thetha = 30 degree and on solving for we get u²=2000Height= u²sin²30/2gH=2000/8gH=25

2 years ago
							y = 10t - t2vy = dy/dt = 10- 2tAt maximum height, vy = 0m/sTime taken to reach the maximum height, t = 5 secondsMaximum height, y = 10(5) - (5)2 = 25 m

one year ago
							By differentiating y = 10t - t^2 we get vertical component of velocity V = 10 -2t. At max height. V = 0, thus we get t = 5s.By substituting it at y = 10t -t^2, we get y = 25m

one year ago
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