A projectile is projected with velocity (3i + 4j + 5k) m/s. The point of projection is taken as origin, and east, north, and vertically upward directions are taken as x, y, and z axes respectively. Find the position of the particle at maximum height.

First, break the velocity into components:

• Horizontal components → 3 m/s (x-direction), 4 m/s (y-direction)
• Vertical component → 5 m/s (z-direction)

Step 1: Time to reach maximum height

At maximum height, vertical velocity becomes zero:

t = uz / g = 5 / 10 = 0.5 s

Step 2: Horizontal displacement

x = 3 × 0.5 = 1.5 m

y = 4 × 0.5 = 2 m

Step 3: Vertical displacement

z = ut - (1/2)gt²

z = 5(0.5) - (1/2)(10)(0.5)² = 2.5 - 1.25 = 1.25 m

Final Position = (1.5i + 2j + 1.25k) m

At maximum height, vertical velocity becomes zero while horizontal motion continues.