Projectile from top and bottom of a tower striking the same target simultaneously: Two shots are fired at the same time from the top and the bottom of a vertical tower toward a ground target 1000 m away (horizontally). The elevation angles are 30° from the top and 45° from the bottom. If both shots hit the target simultaneously, what is the height of the tower?

Introduction / Context:
This kinematics problem links horizontal range relations with vertical motion for two projectiles launched simultaneously from different elevations. It tests comfort with time-of-flight equality and trigonometric components of motion under uniform gravity.

Given Data / Assumptions:

• Horizontal distance to target = 1000 m.
• Top shot elevation = 30° above horizontal; bottom shot elevation = 45°.
• Both hit the same ground target at the same time.
• Neglect air resistance; take g ≈ 9.8 m/s^2.

Concept / Approach:

Let common flight time be t. For the bottom (ground) shot at 45°, use vertical motion to get t, then use it in the top-shot vertical displacement to find tower height H. Horizontal components provide speeds in terms of t but cancel neatly.

Step-by-Step Solution:

Verification / Alternative check:

Reversing the calculation with H = 425 m yields consistent times for both shots to reach the target.

Why Other Options Are Wrong:

350 m, 375 m, and 400 m underestimate the required drop from the top shot; 450 m slightly overestimates compared to the computed 422–423 m.

Common Pitfalls:

Using equal speeds for both shots (not required); forgetting that equal arrival time fixes t via the 45° ground shot; sign mistakes in vertical displacement from the top.

Final Answer:

425 m