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        A boy is running along positive x axis with 8m/s . While running he manages to throw a stone in a plane perpendicular to his direction of running with velocity 12m/s at an angle 30 degree with vertical. Find the speed of the stone at the highest point of the trajectory .
one year ago

Payal
17 Points
							At the Highest point of the trajectory, vertical component of velocity  becomes zero but the horizontal component remains constant i.e. 12sin30 i.e 6m/s..

one year ago
9 Points
							As Cos 60 degree is half so 12cos60 is 6 and it has already speed of 8 metre per second in x-axis hence 8 + 6 is 14 correct answer should be 14

one year ago
Payal
13 Points
							The boy thrw the ball at 12m/s wrt to ground its not mentioned anywhere that h

one year ago
Rishabh Sankla
33 Points
							It is given that the velocity of the boy is 8 m/s along the positive x-axis and the stone is thrown with velocity 12 m/s at an angle of 90° with the direction in which the boy is running. It means the boy would have thrown the stone either in the x-y plane or the x-y´ plane. We know that the vertical velocity of a body at the highest point of its trajectory is zero, so the ball will possess only horizontal velocity at this point. Since the ball is making 60° with the x-axis, its horizontal velocity is given by, $12\frac{m}{s}\cos 60\degree=6\frac{m}{s}$ Since the angle between the direction of the projection of the ball and the direction of the running boy is 90°, the resultant velocity can be calculated by applying Pythagoras’ Theorem. By applying Pythagoras’ Theorem we get $\textrm{Resultant velocity}=\sqrt{(8\frac{m}{s})^2+(6\frac{m}{s})^2}$ $=\sqrt{64\frac{m^2}{s^2}+36\frac{m^2}{s^2}}=\sqrt{100\frac{m^2}{s^2}}=10\frac{m}{s}$

2 months 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