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zheka24 [161]
4 years ago
9

As the diver jumps, the board bends down. What kind of the energy does the board now have?

Physics
1 answer:
Vlada [557]4 years ago
8 0
Hi there!

The energy that is about to or can be released from an object after energy has been transferred to it is called potential energy. In this case, as the diver jumps, potential energy is stored in the board as it bends. The potential energy is then released as kinetic energy when the board springs back up and the diver actually jumps. 

Hope this helps!
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Answer:

(a) The two balls collide 2\; \rm s after launch.

(b) The height of the collision is 4\; \rm m.

(Assuming that air resistance is negligible.)

Explanation:

Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.

The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of g should be negative. The question states that the magnitude of g\! is 10\; \rm m \cdot s^{-2}. Hence, the signed value of \! g should be \left(-10\; \rm m \cdot s^{-2}\right).

Similarly, the initial velocity of the ball thrown downwards should also be negative: \left(-8.0\; \rm m \cdot s^{-1}\right).

On the other hand, the initial velocity of the ball thrown upwards should be positive: \left(12\; \rm m \cdot s^{-1}\right).

Let v_0 and h_0 denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time t:

\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0.

For both balls, g = \left(-10\; \rm m \cdot s^{-2}\right).

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\displaystyle  h(t) = -5\, t^{2} + (-8.0)\, t + 40 (where h is in meters and t is in seconds.)

Similarly, for the ball thrown upwards:

  • Initial velocity: v_0 = \left(12\; \rm m \cdot s^{-1}\right).
  • Initial height: h_0 = 0\; \rm m.

\displaystyle  h(t) = -5\, t^{2} + 12\, t (where h is in meters and t is in seconds.)

Equate the two expressions and solve for t:

-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t.

t = 2.

Therefore, the collision takes place 2\, \rm s after launch.

Substitute t = 2 into either of the two original expressions to find the height of collision:

h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m.

In other words, the two balls collide when their height was 4\; \rm m.

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