Projectile speed at a given height (energy method): A projectile is launched with speed 100.3 m/s at 60°. Neglecting air resistance, what is its speed when it is passing a height of 100 m above the launch level?

Introduction / Context:
Projectile motion with no air resistance conserves mechanical energy when measured relative to a fixed datum. Therefore, the speed at a given height depends only on the initial speed and the gain in potential energy, not on the firing angle (except for feasibility of reaching that height).

Given Data / Assumptions:

• Initial speed u = 100.3 m/s; height h = 100 m.
• Gravity g ≈ 9.81 m/s^2; air resistance neglected.
• Launch and observation height measured from same level.

Concept / Approach:
Use energy: 0.5 * m * u^2 = 0.5 * m * v^2 + m * g * h. Mass cancels. Solve for v. The direction components do not matter for the magnitude of speed when using energy balance.

Step-by-Step Solution:

Verification / Alternative check:
Angle independence: any angle that reaches 100 m yields the same speed magnitude at that height (ignoring drag). A quick check with u = 100 m/s would give v ≈ √(10000 − 1962) ≈ 89.4 m/s, consistent with the above.

Why Other Options Are Wrong:

• 70–85 m/s: underestimate energy remaining after gaining 100 m of height.
• Values above 90 m/s would require less height or higher initial speed.

Common Pitfalls:
Trying to split into horizontal and vertical components and then recombining—valid but longer—while energy is faster and less error-prone.

Final Answer:
90 m/s