**Projectile Launch Speed Calculation****Problem:**The highest barrier that a projectile can clear is 23.1 m, when the projectile is launched at an angle of 12° above the horizontal. What is the projectile launch speed?_Number: _______ units_**Explanation:**We have a given angle of launch and the maximum height that the projectile can reach. To find the launch speed, we can use the following kinematic equations and concepts from projectile motion. The highest point in the projectile's trajectory corresponds to the maximum vertical displacement.Key ideas:- \( \theta = 12^\circ \) (launch angle)- \( h_{max} = 23.1 \) m (maximum height)The vertical component of the launch speed \( v_y \) can be found using the equation for vertical motion:\[ h_{max} = \frac{v_{y}^2}{2g} \]where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.Solving for \( v_y \):\[ v_y = \sqrt{2gh_{max}} \]\[ v_y = \sqrt{2 \times 9.8 \times 23.1} \]\[ v_y \approx \sqrt{452.76} \]\[ v_y \approx 21.27 \, \text{m/s} \]The vertical component \( v_y \) is related to the initial launch speed \( v_0 \) by the equation:\[ v_y = v_0 \sin(\theta) \]Therefore,\[ v_0 = \frac{v_y}{\sin(\theta)} \]Substituting the known values:\[ v_0 = \frac{21.27}{\sin(12^\circ)} \]Finally, calculate \( v_0 \):\[ v_0 \approx \frac{21.27}{0.2079} \]\[ v_0 \approx 102.3 \, \text{m/s} \]So, the launch speed \( v_0 \approx 102.3 \, \text{m/s} \).

Name: **Projectile Launch Speed Calculation****Problem:**The highest barrie
Uploaded: 2024-03-20T06:57:47.000Z
Channel: Bartleby
Description: **Projectile Launch Speed Calculation****Problem:**The highest barrier that a projectile can clear is 23.1 m, when the projectile is launched at an angle of 12° above the horizontal. What is the projectile launch speed?_Number: _______ units_**Expla

Given data: Maximum height (Hmax) = 23.1 m Projectile angle (θ) = 12°, above horizontal Projectile…

Answered: The highest barrier that a progechle…

Science

Physics

The highest barrier that a progechle can clear is 23.1m, When the projectile is launched at an cungle of i2 above horizuntal, what is the prosectile launch speed ne number units

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)...

See similar textbooks

Similar questions

I. If s’(t) = v(t), and s’’(t) = a(t), use the fact that integration reverses differentiation (+C) and the given information about s(t) to find the particular solution to the following problem. If a model rocket’s velocity is given by a(t)= -32 m/s2 and v(0)= 10 and position at 1 second is 1,000 m, find s(t). Hint: Integrate a(t) to find v(t), then integratev(t) to find s(t), finding each C using the given info.
I need the answer quickly
SOLVE STEP BY STEP IN DIGITAL FORMAT In the following problem, break down the procedure by placing: Data, Formula / Clearances, Operations and Result If Captain Marvel throws Spiderman against Thanos to remove his gauntlet, with a throw of 800 meters, with what speed should the tour start? Consider that the Captain uses a 50º angle, measured with her digital protractor. What was her flight time and what maximum height did she reach in her trajectory?
a rocket is fired vertically upwards from ground level with an intial velocity of v0 ft/sec. assuming the rocket obeys Galileo;s law of motion due to gravity, determine the initial velocity required in order for the rocket to attain a height of 1 mile?
I need the answer as soon as possible
Solve the problem. PROVIDE THE GIVEN, REQUIRED, EQUATION, SOLUTION, AND FINAL ANSWER Johannah throws a ball from the window 9.5 m above the ground. When the ball leaves his hand, it is moving at 9.0 m/s at an angle of 25° below the horizontal. How far horizontally from the window will the ball hit the ground? (Ignore air resistance.)
Prove the well-known result that, for a given launch speed vo, the launch angle 0 = 45° yields the maximum horizontal range R. Determine the maximum range if vo = 25 m/s. (Note that this result does not hold when aerodynamic drag is included in the analysis.) Answer: R = i m
Check Your Understanding If the two golf shots in Example 4.9 were bunched at the same speed, which shot would have the greatest range? When we speak of the range of a projectile on level ground, we assume R is very small compared with the circumference of Earth. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given earlier because the projectile has father to fail than it would on level ground, as shown in Figure 4.17, which is based on a drawing in Newton's Principia. If the initial speed is great enough, the projectile goes into orbit. Earth's surface drops every 8000 m. In 1 s an object falls 5 m without a resistance. Thus, if an object is given a horizontal velocity of 8000 m/s (or 18,000 mi/hr) near Earth's surface, it will go into orbit around the planet because, the surface continuously falls away from object Thu is…
If the BB doesn't exit the box, what will the velocity of the box v2, be in meters per second?
. The rocket is fired 200 m downfield from an observer at O. When the rocket has climbed to 100 m in height at P, is has a purely vertical velocity of v = 15 m/s. What is r and 8? 1 200 m 100 m