Q BgQuestion:

      
Scholar
Karma Points: 368
Respect (100%):
posted by  Jim Paulin on 11/3/2009 10:07:18 PM  |  status: Closed  |  Earned Karma: 368

projectiel motion

Course Textbook Chapter Problem Needs by
N/A N/A N/A N/A 11/4/2009 at 7:00:00 AM
Question Details:
How long will it tkae a shell fired from a cliff at an initial velocity of 800 m/s at an angle 30 degrees below the horizontal to reach the ground 150 meters below?  the answer is 0.37 seconds
Tags: Physics
Bonus Point Alert! Earn +4 additional karma points for helping this gold member.

AAnswers:

Answer Question Ask for clarification
posted by esther_jyothi on 11/3/2009 10:10:56 PM  |  status: Live
Asker's Rating: Somewhat Helpful   
Response Details:
Given  
        Horizontal range  R  = 150 m 
                   initial velocity  u  = 800 m/s 
                    angle     θ       =  30 deg 
            we know that    R  = ( u sin θ  )  t 
                                  t   = R  /  u sin θ 
                                        = 150  / 800 x sin 30 
                                         =   0.375 sec
posted by Jim Paulin on 11/3/2009 10:15:08 PM  |  status: Live
Asker's Rating: N/A-Posted by Person Asking Question   
Response Details:
but the horizontal range is not 150 and  is not given, only the vertical distance of 150 meters is known, doesn't acceleration due to gravity have to factor in?
Oracle
Karma Points: 18,830
posted by Marth on 11/3/2009 10:21:39 PM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
The initial velocity is found by .





Use the Quadratic Formula to solve.  




Thank you for rating my response. Feel free to PM me if you have further questions.

Java Genius

"Lois Lane is falling, accelerating at an initial rate of 32ft per second per second. Superman swoops down to save her by reaching out two arms of steel. Ms. Lane, who is now traveling at approximately 120 miles per hour, hits them, and is immediately sliced into three equal pieces." - Sheldon
Novice
Karma Points: 42
posted by fnkyfrsh on 11/3/2009 10:39:42 PM  |  status: Live
Asker's Rating: Helpful   
Response Details:
since the vector here is going down and to the right you will need to use Vcosθ and -Vsinθ
set up the data table
= 800cos(30)  = -800sin(30) or -400m/s
= 800cos(30)   = ?
= ?                  = 0m
= 0m              = 150m       
 a =  0m/s^2            a =  -9.8m/s^2
 t = ?                       t = ?
using the y's
-800sin(30) = -400
= -400^2 + 2(-9.8)(-150)
=
 v = 403.66 but since it is in the down direction it is -403.66
now use
t=
t=
t
t=.37s
Answer Question Ask for clarificarion

Join Cramster's Community

Cramster.com brings together students, educators and subject enthusiasts in an online study community. With around-the-clock expert help and a community of over 100,000 knowledgeable members, you can find the help you need, whenever you need it. Join for free today » How Cramster is different from tutoring »