A Frisbee is lodged in a tree branch 7.0m above the ground. A rock thrown from below must be going at least 3.0m/s to dislodge the Frisbee. How fast must such a rock be thrown upward if it leaves the thrower's hand 1.3m above the ground? Express your answer using two significant figures.

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter2: Motion In One Dimension
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A Frisbee is lodged in a tree branch 7.0m above the ground. A rock thrown from below must be going at least 3.0m/s to dislodge the Frisbee. How fast must such a rock be thrown upward if it leaves the thrower's hand 1.3m above the ground? Express your answer using two significant figures.

Expert Solution
Step 1

Given:

  • The distance of frisbee above the ground is s=7 m.
  • The speed of the rock is v=3 m/s.
  • The height at which rock leaves the thrower's hand h=1.3 m.

 

The expression from equation of motion is given by,

v=u-gtt=u-vg

Here, is the required speed of the rock, is the gravitational acceleration and t is the time when rock touches the frisbee.

 

The expression from equation of motion is given by,

s=h+ut-12gt2

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