A projectile is projected with3i^+4j^+5k^ metre per second point of pr...
Displacement when it hits the xy plane:
- The velocity of the projectile can be broken down into its x, y, and z components: 3i^+4j^+5k^
- When the projectile hits the xy plane, its z component becomes 0 since it is in the xy plane.
- Therefore, the displacement at this point is equal to the x and y components of the velocity, which is 3i^+4j^.
Location of particle at maximum height:
- At the maximum height of the projectile, the vertical component of the velocity becomes 0 since the projectile momentarily stops moving vertically.
- Using the equation of motion, v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is acceleration, and s is displacement, we can find the displacement in the z direction.
- Substituting the values, we get 0^2 = 5^2 + 2*(-10)*s, which gives us s = 2.5 meters.
- Therefore, the location of the particle at its maximum height is 1/2(3i^+4j^+2.5k^).
By following the above steps, we can calculate the displacement when the projectile hits the xy plane and the location of the particle at its maximum height accurately.