
#1
May411, 11:52 AM

P: 22

Hi, im trying to find the initial velocity of this question:
Using a video camera, it is observed that when a ball is kicked from A, it just clears the top of a wall at B as it reaches its maximum height. Knowing the distance from A to the wall is 20m and the wall is 4m high, determine the initial velocity at which the ball was kicked. Neglect size of the ball ( Ans Ua = 23.9m/s). So im thinking to find the angle you have to use trig : so 20^2 + 4^2 = square root of 416 = 20.4m (hypotenuse) using Sin, 4/ 20.4 = Invert Sin 0.196 = 11.3 degrees angle the ball is projected. Now i have then angle what else do i need to find in order to find the initial velocity at which the ball was kicked? Thanks for any help 



#2
May411, 05:29 PM

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P: 40,874

Analyze the vertical motion first. Figure out the vertical component of the initial velocity. 



#3
May411, 06:30 PM

P: 22

Ah ok, i thought about the ball not traveling in a straight line assuming the angle was wrong at some point.
Ok so if 'm trying to analyze the vertical component of the initial velocity im going to use equation: Sy = Uyt + 0.5(ay)t^2 Or Vy^2 = Uy^2  2g(Sy) The 2nd equation i'm thinking i should use since it has just one unknown (Uy). Vy = 0, Sy = 4m 0 = Uy^2  2(9.81)(4) I dont want to go further until i know what i'm doing so far is correct.. 



#4
May411, 06:46 PM

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P: 40,874

Projectile Motion (Initial Velocity) 



#5
May411, 07:21 PM

P: 22

Ok so
Uy^2 = 2(9.81)(4) + 0 Uy^2 = square root 78.48 Uy =8.86 m/s Correct? What shall i do now? Do i find the time? Sy = Uyt + 0.5(9.81)t^ Sy = 4 + (8.86)t  4.905t^2 Sy = 12.86t / 4.905^2 4 = 12.86  4.905 4 = 7.955 t = square root 0.5 = 0.7s t = 0.7 s .. is this correct so far? 



#6
May411, 07:37 PM

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P: 40,874

(1) 4 is the final position, not the initial. (2) a + bt ≠ (a + b)t Even easier: What's the average vertical speed? Use that to find the time. 



#7
May411, 07:53 PM

P: 22

Average vertical speed..
Uy / g 8.86 / 9.81 = 0.9 s? (Sorry im getting tired as its nearly 2am) 



#8
May411, 07:55 PM

P: 22

speed is calculating in m/s.. sorry im confused




#9
May511, 01:36 PM

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P: 40,874





#10
May511, 02:01 PM

P: 22

sice it highest point is Sy = 4m
is it 8.86 / 4 = 2.215 m/s average vertical speed? if its not can you provide further help? thanks 



#11
May511, 02:18 PM

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P: 40,874

You may want to review some of the basic formulas of kinematics: Basic Equations of 1D Kinematics Kinematic Equations and ProblemSolving And here's some good stuff about projectile motion: Projectile Motion 



#12
May511, 02:36 PM

P: 22

u+v / 2 = the average?
So v = 0 8.86 / 2 = 4.43 m/s 



#13
May511, 02:37 PM

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P: 40,874





#14
May511, 02:43 PM

P: 22

with the analogy:
Say you made 0$ to 100$ over 7 days .. average income per day is 100 / 7 = 14.3$ per day 



#15
May511, 02:57 PM

P: 22

Sy = (4.43)t  (4.91)t^2
4 = (4.43)t  (4.91) 4 = 0.48? 4 / 0.48 = square root 8.33 t = 2.887 s ? 



#16
May511, 02:58 PM

P: 22

Question: Why does the AVERAGE Initial vertical velocity have to be used instead of just the initial vertical velocity so i can understand why?




#17
May511, 03:21 PM

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P: 40,874

0/day; 1/day, 2/day, 3/day...... 100/day. Average (assuming uniform increasejust like uniform acceleration) is (0 + 100)/2 = 50/day. Half the time you made less, half the time you made more. (But don't worry about that analogy if it doesn't help.) 



#18
May511, 03:26 PM

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P: 40,874

I wanted to show you an easier way to get the time, using Distance = ave speed X time. No quadratics to solve. (Do it both ways and compare the answers.) 


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