Maximum height of a projectile (given u = 4 m/s): A particle is projected from a horizontal plane with initial speed 4 m/s at an angle α above the horizontal. What is the maximum height reached, expressed in terms of α and g?

Introduction / Context:
Maximum height depends on the initial vertical component of velocity and gravity. With a fixed launch speed u, the height varies with sin^2 α because vertical kinetic energy converts to gravitational potential energy at the apex where vertical speed becomes zero.

Given Data / Assumptions:

• Initial speed u = 4 m/s.
• Projection angle = α (above horizontal).
• Uniform gravity g; no air resistance.

Concept / Approach:

Standard formula for maximum height on level ground: H = (u^2 sin^2 α) / (2 g). Substituting u = 4 gives a simplified expression.

Step-by-Step Solution:

Verification / Alternative check:

Energy method: (1/2) m (u sin α)^2 converts to m g H → H = u^2 sin^2 α / (2 g), confirming the kinematic derivation.

Why Other Options Are Wrong:

(b) involves sin 2α (related to range, not height); (c) misses the factor 2 in the denominator after substitution; (d)–(e) use cos^2 α which pertains to horizontal component, not maximum height.

Common Pitfalls:

Confusing maximum height with range expressions; forgetting to square the sine of the launch angle; arithmetic slips when substituting u = 4 m/s.

Final Answer:

(8 / g) sin^2 α