Question

A ball is thrown upward with a speed of 30 m/s. Approximately how much time does it take the ball to travel from the release location (A) to its highest point (B)? Approximately how much total time is the ball in the air before it returns back to its original height (C)?

Answer 1

Answer:

The answer is below

Explanation:

Using the equation of newton law of motion for an object thrown vertially upwards:

v = u - gt

where u is the initial velocity, g = acceleration due to gravity and t is the time taken to reach maximum height.

Given that u = 30 m/s, at maximum height, v = 0, g = 10 m/s². Therefore:

0 == 30 - 10t

10t = 30

t = 30/10

t = 3 seconds

b) The time taken to reach the highest point = 3 seconds. Therefore the total time before it returns back to its original height = 2t = 2(3) = 6 seconds