badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 11

                        

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

Answers : (2)

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 problems
Always 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 = 250
where v_{x} = velocity in x direction
t = total time
We also know that time taken for projectile to reach maximum height will be half of total time. so t/2
Now 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 answer
Now , we know the maximum time will be when projectile takes maximum height
Or when x component of velocity is 0.
We need to find half of that time.
For maximum height,
v_{y} = gt
t = 5 secs
Now 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 = 25
Solving 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
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

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


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details