solve Q 9

solve Q 9 wnerev IS In and y isin m. It is also known that x = O when t = 0. Determine (i) Acceleration of particle (ii) Velocity of particle when y 5. The position of a particle is given by x = 7 + 3t3 m and y = 13 + 5t —9t2 m, where x and y are the position coordinates, and t is the time in s. Find the speed (magnitude of the velocity) when the x component of the acceleration is 36 m/s2. Projectile motion 6. 7. 8. 9. 10. A particle is projected with a speed of 10 rn/s at an angle 370 with the vertical. Find (i) time of flight (ii) maximum height above ground (iii) horizontal range. A particle is thrown with a speed 60 ms-l at an angle 600 to the horizontal. When the particle makes an angle 300 with the horizontal in downward direction, it's speed at that instant is v. What is the value ofv2 in SI units ? A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ? A particle is projected upwards with a velocity of 100 m/s at an angle of 600 with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking g m/s2. A panicle is projected in x-y plane with y-axis along vertical, the point of projection is origin. The 2 equation of a path is y = &fix . Find angle of projection and speed of projection. 13

Dear Student

Let’s assume that initial velocity of the particle is
 u = u cosθ i + u sin θ j
Here, positive j axis is taken as vertically upward.
In projectile motion, vertical component of velocity changes but horizontal component of velocity remains always constant.
Therefore let at point where
particle will move perpendicular to its initial direction velocity is v
v = u cosθ i + (u sin θ -gt) j   (As vertical component changes according to kinematics equation v = u -gt)

Now as at this point as particle is moving perpendicular to initial direction so
u.v =0u cosθi^+u sinθj^.u cosθi^+(u sinθ-gt)j^=0u2cos2θ+u2sin2θ-u sin θgt=0t=ug sinθ=10010 sin60=203


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