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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

Kshitij Sharma
33 Points
							D.25m is the correct answer.Using vector equation $\vec{r}=\vec{u}t+\frac{1}{2}\vec{{g}'}t^{2}$​ where ${g}'$​ is acc. due to grav. on the planet, we get :  $\vec{r}=u\cos \theta t\hat{i}+(u\sin \theta t-{g}'t^{2})\hat{j}$​ = $10\sqrt{3}t\hat{i}+(10t-t^{2})\hat{j}$​ . On comparing and evaluating corresponding terms, we get $\tan \theta =\frac{1}{\sqrt{3}}$   so,  $\theta =30^{\circ}$​ , |u|=20m/s and $|{g}'|=2ms^{-2}$ .From $y_{max}=\frac{u^{2}\sin ^{2}\theta }{2{g}'}$​, max. height=25 m.

2 years ago
utpal raj
244 Points
							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
Gaurav
13 Points
							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
Animesh Agarwal
11 Points
							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
Jayakeerti
17 Points
							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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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions