A stone is thrown from the top of a building upward at an angle of 26.0° to the horizontal with an initial speed of 21.0 m/s as shown in the figure. The height from which the stone is thrown is 45.0 m above the ground. (0, 0) 45.0 m y-45.0 m A stone is thrown from the top of a building.
A stone is thrown from the top of a building upward at an angle of 26.0° to the horizontal with an initial speed of 21.0 m/s as shown in the figure. The height from which the stone is thrown is 45.0 m above the ground. (0, 0) 45.0 m y-45.0 m A stone is thrown from the top of a building.
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
Chapter3: Motion In Two Dimensions
Section: Chapter Questions
Problem 21P: A playground is on the flat roof of a city school, 6.00 m above the street below (Fig. P3.21). The...
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Express the vertical position of the stone from the particle under constant acceleration model (Use the following as necessary: vyi, g, and t. Do not substitute numerical values; use variables only.):
yf = yi + _ __
-45.0m = 0 + (9.21 m/s)t + 1/2 ( -9.80)t^2
Solve the quadratic equation to calculate t (in s):
t =
Use the velocity equation in the particle under constant acceleration model to obtain the y-component of the velocity of the stone just before it strikes the ground (Use the following as necessary: vyi, g, and t. Do not substitute numerical values; use variables only.):
vyf = vyi − _________
Substitute numerical values (Enter your answer in m/s.):
vyf =
Use this component with the horizontal component vxf = vxi = 18.9 m/s to find the speed of the stone (in m/s) at t:
vf = sqrt(vxf2 + vyf2
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