In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-like nature of the board to help propel him into the air. Assume that the diver's motion is essentially vertical. He leaves the board, which is 3.0 m above the water, with a speed of 6.3 m/s. The diver is in the air for 1.7 s from when he left the board to the water. What is the speed of the diver when he reaches the water?
Express your answer with the appropriate units.

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skyluke89 skyluke89 · Answer 1 · 2019-10-08T10:47:55+02:00

Answer:

10.4 m/s

Explanation:

The problem can be solved by using the following SUVAT equation:

[tex]v=u+at[/tex]

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the diver in the problem, we have:

[tex]u=+6.3 m/s[/tex] is the initial velocity (positive because it is upward)

[tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity (negative because it is downward)

By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

[tex]v=+6.3 + (-9.8)(1.7)=-10.4 m/s[/tex]

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.

aristocles aristocles · Answer 2 · 2019-10-23T03:36:02+02:00

Answer:

final speed with which it will hit the water is v = 9.92 m/s

Explanation:

Here we know that diver's motion is essentially in air

So the motion of diver is under gravity at uniform acceleration

so we can use kinematic in order to find the final velocity

so we will have

[tex]v_f^2 - v_i^2 = 2ad[/tex]

[tex]v_f^2 - 6.3^2 = 2(9.8)(3)[/tex]

so we will have

[tex]v_f = 9.92 m/s[/tex]

so final speed with which it will hit the water is v = 9.92 m/s