a body is projected horizonatlly from the top of a cliff 45m above the ground. if the body lands at a distance…

Analyze the question: A body is projected horizontally from a cliff 45 m high and lands 30 m from the foot of the cliff. We need the horizontal speed at launch.

Concept: In projectile motion, the horizontal and vertical motions are independent.

• Vertical: the initial vertical velocity is 0, so h = ½gt²
• Horizontal: there is no acceleration, so range R = ut

Solution:

Step 1: Find the time of flight from the vertical motion.

45 = ½ × 10 × t² 45 = 5t² t² = 9 t = 3 s

Step 2: Use the horizontal motion to find the speed of projection.

R = ut 30 = u × 3 u = 10 m/s

Answer: The speed of projection is 10 m/s, which is option A.

Tip: Always find the time from the vertical motion first, because time is the link between the two motions. For horizontal projection, t = √(2h/g), and then u = R/t.