Help with # 5 bars-epresenting xE Keandra Nabors-Horizontally la xdocs.google.com/document/d/110BCZol7VTQ97k78khcVt4QwmCiLrh.JYIC92NC69YWQ/editNabors- Horizontally launched projectiles D OView Insert Format Tools Add-ons HelpLast edit was seconds ago100%Normal text - ArlalBIUA8田回,一明 |三|E▼三114.12.31.3.5. Will smith pushes the irobot over the ledge to its doom. If the robot had a horizontalvelocity of 2.3 m/s and was 17 m above ground. What was the robots horizontaldisplacement? When it hit the ground?Horizontally launched projectilesPart B Instructions: Use the information above to solve each of the following problems. CheckYOur answers helow, Attach your work to the d0cument

Answered: 5. Will smith pushes the irobot over…

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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Help with # 5

bars-epresenting x
E Keandra Nabors-Horizontally la x
docs.google.com/document/d/110BCZol7VTQ97k78khcVt4QwmCiLrh.JYIC92NC69YWQ/edit
Nabors- Horizontally launched projectiles D O
View Insert Format Tools Add-ons Help
Last edit was seconds ago
100%
Normal text - Arlal
BIUA
8田回,一明 |三|E▼三
11
4.
1
2.
3
1.
3.
5. Will smith pushes the irobot over the ledge to its doom. If the robot had a horizontal
velocity of 2.3 m/s and was 17 m above ground. What was the robots horizontal
displacement? When it hit the ground?
Horizontally launched projectiles
Part B Instructions: Use the information above to solve each of the following problems. Check
YOur answers helow, Attach your work to the d0cument

Expert Solution

Step 1

Given

The horizontal velocity is u_h = 2.3 m/s.

The height from which robot is pushes or the vertical distance cover by the robot is h = 17 m.

By the newton equation of motion, the vertical distance formula is given as,

$h = u_{v} \times t + \frac{1}{2} g t^{2}$

Here, u_v is the vertical initial velocity which is equal to 0 as it is free fall from the height, t is the time it hits the ground and g is acceleration due to gravity whose value is 9.8 m/s².

Substitute the known values.

$\begin{array}{rcl} 17 m & = & 0 \times t + \frac{1}{2} \times 9.8 m / s^{2} \times t^{2} \\ t^{2} & = & \frac{17 m \times 2}{9.8 m / s^{2}} \\ = & 3.4694 s^{2} \\ t & = & \sqrt{3.4694 s^{2}} \\ = & 1.86 s \end{array}$

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