A diver jumps off a 10m diving board with an initial upward velocity of 5.0 m/s. What is the diver's velocity after 1.5 seconds?
Initially, the diver is moving upwards
0 = Vo – gt,
t = Vo/g ≈ 5/10 = 0.5 s.
After 0.5 seconds of motion the diver is at the maximum height.
Then the downward motion begins.
After 1.5 - 0,5 = 1 second of motion the velocity of diver is
v = g•t = 10•1= 10 m/s.
To determine the diver's velocity after 1.5 seconds, we can use the laws of motion. The initial velocity is 5.0 m/s in the upward direction, which means the velocity has a positive value.
In this case, the downward acceleration due to gravity will cause the diver's velocity to decrease over time. The acceleration due to gravity is approximately 9.8 m/s², acting in the downward direction.
Using the equation of motion:
v = u + at
where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time.
Given:
u = 5.0 m/s (upward)
a = -9.8 m/s² (downward, negative because it opposes the initial motion)
t = 1.5 seconds
Substituting the values into the equation:
v = 5.0 m/s + (-9.8 m/s²) * 1.5 s
Calculating:
v = 5.0 m/s - 14.7 m/s
The diver's velocity after 1.5 seconds is approximately -9.7 m/s.
The negative sign indicates that the velocity is directed downward, opposite to the initial upward direction.