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Chapter 2: Problem 41

A Frisbee is lodged in a tree 6.5 m above the ground. A rock thrown from below must be going at least \(3 \mathrm{m} / \mathrm{s}\) to dislodge the Frisbee. How fast must such a rock be thrown upward if it leaves the thrower's hand \(1.3 \mathrm{m}\) above the ground?

Short Answer

Expert verified
The rock must be thrown upwards with an initial velocity of approximately \( 10.20 \mathrm{m} / \mathrm{s} \) to dislodge the Frisbee.

Step by step solution

01

Determine the height to reach

Calculate the height the rock needs to reach by subtracting the height at which the rock is being thrown from the total height of the Frisbee above the ground: \( h = 6.5 m - 1.3 m = 5.2 m \).
02

Set up and solve the motion equation

We will use the following motion equation: \( h = v_i*t + 0.5*a*t^2 \). However, we don't know the time, t. A trick is to use another motion equation, specifically the one that does not contain the time: \( h = v_i^2/(2a) \). Now we can solve this equation for \( v_i \), the initial velocity, using \( h = 5.2 m \) and \( a = -9.8 m/s^2 \):\n\n\( 5.2 = v_i^2 / (2*-9.8) \)\n\n\( v_i = sqrt(5.2 * 2 * 9.8) \)
03

Calculate the initial velocity

Insert the values of h, a in the equation solved in step 2 and compute the value of \( v_i \):\n\n\( v_i = sqrt(5.2 * 2 * 9.8) = 10.20 m/s \)

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Most popular questions from this chapter

Starting from home, you bicycle 24 km north in 2.5 h and then turn around and pedal straight home in \(1.5 \mathrm{h}\). What are your (a) displacement at the end of the first \(2.5 \mathrm{h},\) (b) average velocity over the first \(2.5 \mathrm{h},\) (c) average velocity for the homeward leg of the trip, (d) displacement for the entire trip, and (e) average velocity for the entire trip?

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