QUESTION 3 A juggler throws a bowling pin straight up with an initial speed of 6.7 m/s from an initial height of 3.6 m. How much time elapses until the bowling pin returns to the same initial height?

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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QUESTION 3
A juggler throws a bowling pin straight up with an initial speed of 6.7 m/s from an initial height of 3.6 m. How much time
elapses until the bowling pin returns to the same initial height?
QUESTION 4
Problem
In a relief operation mission involving a helicopter h distance above the ground traveling with a horizontal velocity vx. What
should be the expression for the horizontal distance dy at which you release the relief package so that it will arrive to the
survivors at the right place? (Neglect the effect of air resistance)
Solution
To determine this, we must first derive the time it takes for the relief package to reach the survivors. We use this equation
-h = Vinitial-y' + (1/2)ay
If we just drop the package from the helicopter, the equation above becomes
+ (1/2)ay
%3D
Substituting ay = -g then simplifying results to
t = sqrt(
!3!
which is the time it takes for the object to reach the ground.
Since the package will just travel at a constant velocity in the x-axis, thus
dx = Vxt
%3!
Substituting the time taken by the package to reach to ground results to:
dx = (
( sqrt (
which is the expression for the horizontal distance at which you should drop the package.
Transcribed Image Text:QUESTION 3 A juggler throws a bowling pin straight up with an initial speed of 6.7 m/s from an initial height of 3.6 m. How much time elapses until the bowling pin returns to the same initial height? QUESTION 4 Problem In a relief operation mission involving a helicopter h distance above the ground traveling with a horizontal velocity vx. What should be the expression for the horizontal distance dy at which you release the relief package so that it will arrive to the survivors at the right place? (Neglect the effect of air resistance) Solution To determine this, we must first derive the time it takes for the relief package to reach the survivors. We use this equation -h = Vinitial-y' + (1/2)ay If we just drop the package from the helicopter, the equation above becomes + (1/2)ay %3D Substituting ay = -g then simplifying results to t = sqrt( !3! which is the time it takes for the object to reach the ground. Since the package will just travel at a constant velocity in the x-axis, thus dx = Vxt %3! Substituting the time taken by the package to reach to ground results to: dx = ( ( sqrt ( which is the expression for the horizontal distance at which you should drop the package.
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