a boy standing n a high cliff dives into the ocean below and strikes the water after 2.5 s. what is the boy's velocity when he hits the water? deltav=gxt
As you have already stated,
deltav=gxt
That means the change in velocity (delta V) equals the acceleration of gravity (g = 9.8 m/s^2) times the time spent falling (2.5 s)
To calculate the boy's velocity when he hits the water, we need to use the formula:
Δv = g x t
Where:
Δv is the change in velocity (which represents the final velocity),
g is the acceleration due to gravity, and
t is the time it takes for the boy to hit the water.
In this case, we are given that the time is 2.5 seconds. We can assume that the acceleration due to gravity is approximately 9.8 m/s².
Now, let's calculate the velocity:
Δv = 9.8 m/s² x 2.5 s
Δv = 24.5 m/s
Therefore, the boy's velocity when he hits the water is 24.5 m/s downwards (assuming positive velocity is upward).