Question

A billiard ball leaves a 0.50m high table with an initial velocity of 2.5m/s
a. how long will it take the billiard ball to fall to the ground?
b. what is the horizontal distance between the edge of the table and the landing location of the ball?

Answer 1

Answer:

(a) 0.2s (b) 0.75m

Given; Height of table, 's' = 0.50m

Initial velocity = 2.5m/s

Concept: When object falls downwards it initial velocity is zero. Thus, the final velocity will be 2.5m/s.

Explanation:

a) Speed = Distance/Time

=> v = s/t

=> t = s/v

=> t = 0.50m/2.5m/s

=> t = 0.2s

b) Horizontal distance = ?

Acceleration = (Final-Initial Velocity)/Time

=> a = 2.5m/s-0/0.2s

=> a = 12.5m/s^2

Using 2nd equation of motion;

s = ut+½at^2

=> s = (2.5m/s)(0.2s)+½(12.5m/s^2)(0.2s^2)

=> s = 0.5m+ ½ x 0.5m

=> s = 0.5m + 0.25

=> s = 0.75m

Therefore, horizontal distance = 0.75m

I am not sure this is right because all the values are not provided.