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Please solve the attached questions with steps. Thanks in advance. Can you please guide me how to solve projectile motion problems and circular motion problem accurately
Please solve the attached questions with steps. Thanks in advance. Can you please guide me how to solve projectile motion problems and circular motion problem accurately


4 years ago

komali
19 Points
							the answer is 50m/s and it is surely correct no need to worry about it and proceed please and steps are as follows

4 years ago
Manas Shukla
102 Points
							First of all the answer to the above question is option d and not option c as stated by komali. -.-‘Now,For projectile motion problemsAlways resolve motion in two directions , x and y component.They will become 2x  one-D motion problems.Put the data u get from solving one equation into another and it will become very easy.Lets take the asked question as an example with this method.Taking a look at the graph we see projectile’s maximum range is 250m which occurs at 45 degrees for normal projectile motion with plane surface and constant gravitational field.We know at this angle both x component and y component of velocity are equal during launch.Now lets consider the motion in x direction$Range(R) = v_{x} t$ = 250where $v_{x} =$ velocity in x directiont = total timeWe also know that time taken for projectile to reach maximum height will be half of total time. so t/2Now lets consider motion in y direction while ascending$v_{y} = gt$Substituting value of t from one equation to another we get$250 = \frac{v_{y}v_{x}}{g}$we get $v_{x} = v_{y} = 50$Thus net velocity $v= 50\sqrt{2}$From seeing the options it is very easy to see that the answer will be option d since all of the options exceed or equal 50m/s. But just for sake of explaining and getting the answer lets continue till we get the correct answerNow , we know the maximum time will be when projectile takes maximum heightOr when x component of velocity is 0.We need to find half of that time.For maximum height,$v_{y} = gt$t = 5 secsNow when time of flight is 2.5 secs.$v_{y} = gt$$v_{y} =25$We know that the speed of the projectile will be minimum when y component of velocity becomes 0 thus vx will be our answer.$V = \sqrt{v_{y}^{2} + v_{x}^{2}}$V = 50 as we found out and vy = 25Solving we get $v_{x} = \sqrt{50^{2} - 25^{2}}$$v_{x} =25 \sqrt{4 - 1}$$v_{x} =25 \sqrt{3}$I hope that helps clear your doubts.Keep in mind that the answer got so lengthy cause i was trying to explain. You can literally solve this question very easily when u get used to the method. Approve if that helped , Thanks.

4 years 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