Question

A ball is thrown horizontally from a cliff at a speed of 10 m/s. you predict that its speed 1 s later will be slightly greater than 14 m/s. your friend says it will be 10 m/s. show who is correct.

Answer 1

The horizontal component of speed remains constant at 10 m/s.
After 1 s, a vertical component of speed = gt = 9.8(1) = 9.8 m/s is added to the horizontal speed.
The size of the total speed of the ball after 1 s = √(10²+9.8²) = √196.04 ≈ 14.0 m/s <=
You are correct.

Answer 2

Answer:

Speed is slightly greater than 14 m/s.          

Explanation:

The ball will under go projectile motion. The horizontal velocity remains constant while there would be increase in vertical velocity due to acceleration due to gravity in downward direction.

Using first equation of motion, after 1 s, vertical velocity will be:

v = u +at

v=0+(9.8 m/s²)(1 s) = 9.8 m/s

Horizontal velocity = 10 m/s

Net velocity:

$V= \sqrt{(9.8)^2+(10)^2}=\sqrt{196.04}=14.001 m/s$

Speed is slightly greater than 14 m/s.