Problem 4An object is launched at a velocity of 25 m/s in a direction making an angle of 20° upward with the horizontal.What is the maximum height reached by the object? PHY 150 PROBLEM SET 3Horizontal Velocity:Vertical Velocity:Vertical Velocity at some time t:Vertical Velocity at height y:Horizontal Distance Traveled:Vertical Distance Traveled:Time to Highest Point:Total Flight Time:Projectile Motion Equationsv = v cose = constantX0V = v sine0V = v (sin0) - gty22(v sine)² - 2gAyyAx = v (cos0)tAy = (v sine)t +gi²v sine0t =g2v sine0t=gwhere g=m9. 8-

Name: Problem 4An object is launched at a velocity of 25 m/s in a direction
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Description: Problem 4An object is launched at a velocity of 25 m/s in a direction making an angle of 20° upward with the horizontal.What is the maximum height reached by the object? PHY 150 PROBLEM SET 3Horizontal Velocity:Vertical Velocity:Vertical Velocity at

In the given question, Initial velocity of the projectile (u) = 25m/s Angle of projectile (θ) = 200…

Answered: An object is launched at a velocity of…

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter3: Vectors

Section: Chapter Questions

Problem 3.62AP: After a ball rolls off the edge of a horizontal table at time t = 0, its velocity as a function of...

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Problem 4

An object is launched at a velocity of 25 m/s in a direction making an angle of 20° upward with the horizontal.

What is the maximum height reached by the object?

PHY 150 PROBLEM SET 3
Horizontal Velocity:
Vertical Velocity:
Vertical Velocity at some time t:
Vertical Velocity at height y:
Horizontal Distance Traveled:
Vertical Distance Traveled:
Time to Highest Point:
Total Flight Time:
Projectile Motion Equations
v = v cose = constant
X
0
V = v sine
0
V = v (sin0) - gt
y
2
2
(v sine)² - 2gAy
y
Ax = v (cos0)t
Ay = (v sine)t +gi²
v sine
0
t =
g
2v sine
0
t
=
g
where g
=
m
9. 8-

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