Equal horizontal ranges with different projection angles Two projectiles have projection angles 64° and 45°. If the first has initial speed 10 m/s, what initial speed must the second have to obtain the same horizontal range (neglect air resistance)?

Introduction / Context:
The range on level ground depends on u^2 sin 2θ / g. For equal ranges, speeds and double-angles must satisfy a proportionality. This problem checks manipulation of the range formula and trigonometric identities.

Given Data / Assumptions:

• u1 = 10 m/s at θ1 = 64°.
• u2 = ? at θ2 = 45°.
• Same launch and landing elevations, no air resistance, constant g.

Concept / Approach:

Level-ground range: R = u^2 sin 2θ / g. For R1 = R2, set u1^2 sin 2θ1 = u2^2 sin 2θ2 and solve for u2. Use sin 90° = 1 and the identity sin(180° − x) = sin x.

Step-by-Step Solution:

Verification / Alternative check:

Back-substitute: R ∝ u^2 sin 2θ; with u2 = 8.9 m/s and sin 90° = 1, the ranges match within rounding.

Why Other Options Are Wrong:

7.8 and 6.5 m/s are too small; 9.8 and 10.0 m/s are too large for the given sin 128° factor.

Common Pitfalls:

Using θ instead of 2θ in the sine term; rounding too early leading to larger error.

Final Answer:

8.9 m/s