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Phy-XI-4-7 projectile derivations
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Derivations of different parameters in projectile motion are described in this lesson

State topper in school exams, selected in IIT Kharagpur, M.Sc., 30 yrs in teaching.

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hello ...i think on 5th slide ques no. 3 equation you made is incorrect or not appropriate as it is said A left the work after 4 days so the equation shud be taken as 4(a+b)+18b since A & B both worked for 4 days ..same goes for the second statement ...please clear it..i think the ques is wrong... TIA
Read the question once again. It is clearly mentioned that A started the work, not A and B together, so we cant take 4(A+B) +18B
sir can u pls teach the concepts of inclined projectile and from a height special conditions?
10 months ago
Ok I will include it tomorrow.
ok sir thank you very much
plz complete class 11 physics , as soon as possible
10 months ago
Sure
Little Kumar samal
10 months ago
thank you sir
1. PHYSICS Pradeep Kshetrapal Class XI

2. Chapter 4 Motion in a plain Projectile Motion Lesson No.Phy-XI-4-7

3. The next step of finding relation between quantities in two dimensional motion is to use equations of kinematics and applly them separately for horizontal and vertical direction.

4. In horizontal way the acceleration due to gravity s zero So, distance covered is s- V x t In the horizontal, during the whole flight velocity is same as initial velocity. So, horizontal velocity remains constant through out the flight.

5. In vertical direction the velocity is accelerated velocity. Acceleration ag. The equations in vertical component are VtUSin-gt

6. Different parameters of projectile motion. TIME OF FLIGHT Trajector)y Hmax ucos 0 Range Flight of projectile is continued only till object is above the ground therefore the time of flight depends on vertical motion.

7. But, in horizontal direction we don't know the displacement so, we cant use the formula for displacement-Velocity x time. Remember, Time of flight is always decided by vertical motion. In vertical motion the object reach a certain height and again hits the ground. So the total displacement is zero.

8. By the equation s- ut + at2 2 (In vertical direction.) Initial velocity- usin 0 Acceleration-2. Final vertical Displacement- 0 Time =T.

9. 2 2 Divide both sides with T 1 2u sin Time of flight T-usine

10. Maximum height achieved, Hmax Condition for maximum height is that at that point vertical velocity ecomes zero initial vertical velocity-u sin acceleration- -g final vv-0 S= Hmax = ? max

11. Height is maximum when Sin is maximum or -90. Then Hmax-u?sin290-42 Then Hmax 2g2g

12. Range is maximum when Sin20-1 (Maximum). So, 20-90. or =45. Hence at 45 degree projection the range is maximum This maximum range is 2 max

13. For any given velocity, maximum range is double the maximum height Maximum range ( at 45)- Rmax- ^ At this angle 45 the maximum 2 height is Hmax Sin290 U2 2g