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Erica (39kg \;kg ) and Danny (45kg \;kg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is moving upward past her at 4.8m/s \; m/s . At that instant he grabs hold of her.

What is their speed just after he grabs her?

help please

can you show formula too

Respuesta :

usmanbiu

Answer:

2.57 m/s

Explanation:

mass of Erica (ME) = 39 kg

velocity of Erica (VE) = 0 m/s (since she is at her high point of her jump, her velocity will be 0)

mass of Danny (MD) = 45 kg

velocity of Danny (VD) = 4.8 m/s

from the conservation of momentum total initial momentum is equal to the total final momentum

total initial momentum = final initial momentum

(ME)(VE) + (MD)(VD) = (ME + MD) V

where "V" is the velocity of Erica and Danny after Danny grabs her.

(39 x 0) + (45 x 4.8) = (39 + 45) x V

0 + 216 = 84 V

V = 216 / 84 = 2.57 m/s

Lanuel

The speed just after Danny grabs Erica is equal to 2.57 m/s.

Given the following data:

  • Mass of Erica = 39 kg
  • Mass of Danny = 45 kg
  • Velocity of Erica = 0 m/s (since she is at the highest point)
  • Velocity of Danny = 4.8 m/s

To determine their speed just after Danny grabs Erica, we would apply the law of conservation of momentum:

[tex]M_EV_E + M_DV_D = (M_E + M_D)V_{f}[/tex]

Where:

  • [tex]M_E[/tex] is the mass of Erica.
  • [tex]M_D[/tex] is the mass of Danny.
  • [tex]V_f[/tex] is the final speed.
  • [tex]V_E[/tex] is the speed of the Erica.
  • [tex]V_D[/tex] is the speed of the Danny.

Substituting the given parameters into the formula, we have;

[tex]39 \times 0 + 45 \times 4.8 = (39 +45)V_f\\\\0+216=84V_f\\\\V_f = \frac{216}{84} \\\\V_f = 2.57 \;m/s[/tex]

Final speed = 2.57 m/s

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