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Session Objectives : Session Objectives Equations of Motion
Applications www.aptacads.com
Correct form of Equations of motion : www.aptacads.com Correct form of Equations of motion Position vector of the particle Displacement vector of the particle
Example 1 : www.aptacads.com Example 1
Solution : www.aptacads.com or 4.9t2 – 9.8t – 14.7 = 0 or t2 – 2t – 3 = 0 Þ t = 3 s Solution
Slide 6 : www.aptacads.com vx(t) = 9.8 Ö3 m/s vy(t) = vy(0) + ayt = 9.8 – 9.8 × 3 or vy = –19.6 m/s Solution
Solution : www.aptacads.com Maximum height attained above the ground = Height of B above the point of projection + Height of building Solution
Example 2 : www.aptacads.com Example 2
Solution : www.aptacads.com \ a(2) = –6 × 2 + 2 = –10 ms–2 Solution
Example 3 : www.aptacads.com A mass m is being pulled up using two
strings passing over two pulleys so that
the downward velocities of the strings
are equal to u. (see figure). What is
the speed v of the mass ? Example 3
Solution : www.aptacads.com v cosq = u Solution
Projectile Motion Revisited : www.aptacads.com Projectile Motion Revisited Conclusion:
The x-component of the velocity does not change with time.
In the y-direction, the motion of the projectile is identical to that of a particle thrown vertically upward with velocity u sin q.
Projectile Motion Revisited : www.aptacads.com Projectile Motion Revisited The plot of x and y w.r.t. time is given below.
Projectile Motion : www.aptacads.com Projectile Motion Relation between x and y: This shows that the path of the projectile is parabolic.
Total Time of Flight : www.aptacads.com Total Time of Flight When the projectile gets back to the ground, its y displacement becomes zero. Hence This is the total time of flight.
Horizontal Range : www.aptacads.com Horizontal Range The horizontal distance covered during this time is
Maximum Height : www.aptacads.com Maximum Height The time taken to reach the maximum height is half the total time of flight. Therefore
Example : www.aptacads.com Example
Solution : www.aptacads.com Solution 120 cos60° = v cos45° Hence answer is (a)
Example : www.aptacads.com Example
Solution : www.aptacads.com Solution Comparing the equation with that of the trajectory of a projectile. Hence answer is (a)
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Slide 23 : www.aptacads.com