A coin is pushed from the edge of a laboratory bench with a horizontal velocity of 15.0 ms^ -1 . If the height…

Analyze the question: The coin leaves the bench horizontally, so it is a projectile problem. The horizontal and vertical motions are independent. We need the horizontal distance (range) travelled before it hits the floor.

Concept:

• Vertical motion: initial vertical velocity = 0, so h = ½gt²
• Horizontal motion: constant velocity, so R = ut

Solution:

Step 1: Time of fall

h = ½gt² 1.5 = ½(10)t² 1.5 = 5t² t² = 0.3 t = √0.3 ≈ 0.548 s

Step 2: Horizontal distance

R = ut = 15.0 × 0.548 ≈ 8.22 m

Answer: The distance is about 8.22 m, which is option C.

Tip: For a horizontal launch from a height, always find the time of fall first using h = ½gt² (the horizontal speed does not affect it), then multiply by the horizontal velocity. Also, you can quickly eliminate D (15 m would need a 1 s fall, which would require a 5 m drop) and A and B (too small for such a high speed).