Question

A ball bearing is projected vertically upwards from the ground with a velocity of 15ms. Calculate the time taken by the ball to return to the ground (g=10ms^2)​

Answer 1

Answer:

t = 3 [s]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}=v_{o}-g*t$

where:

Vf = final velocity [m/s]

Vo = initial velocity = 15 [m/s]

g = gravity acceleration = 10 [m/s²]

t = time [s]

Now replacing we have:

$0 = 15 -10*t\\10*t=15\\t= 1.5[s]$

Note: In the equation above the gravity acceleration is negative, because the movement of the ball bearing is pointing againts the gravity acceleration.

The time calculated is only when the ball bearing reaches the highest elevation, and it will take the same time for descending, therefore the total time is:

t = 1.5 + 1.5 = 3 [s]