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Projectile Motion Problem with thrown ball

1. Homework Statement
You are to throw a ball with a speed of 16.5 m/s at a target that is height h = 4.90 m above the level at which you release the ball. You want the ball's velocity to be horizontal at the instant it reaches the target.
- What is the speed of the ball just as it reaches the target?

2. Homework Equations
Voy = Vo sin (θ) = 36.44°.
V^2 = Vo^2 - 2a(∆y)

3. The Attempt at a Solution
I thought Vox = Vx (velocity at the top since Vy = 0) = Vo cos (θ) = 5.0599m/s but that was not right so I tried to solve for t using Vy = Voy +at . . 0 = 16sin(36.44) + (-9.8)t and t came out to be 1.6s so i plugged T=1.6s into x = x0 + Vox(t) . . .13.27 = 0 + Vox(1.6) and got 8.29m/s and that too, is wrong. . . What am missing from me getting the correct answer?
 

Answers and Replies

50
0
1. Homework Statement
You are to throw a ball with a speed of 16.5 m/s at a target that is height h = 4.90 m above the level at which you release the ball. You want the ball's velocity to be horizontal at the instant it reaches the target.
- What is the speed of the ball just as it reaches the target?
That isn't solvable given the data you have there. You need an angle for Vo, or the horizontal distance to the target, or the time.

2. Homework Equations
Voy = Vo sin (θ) = 36.44°.
V^2 = Vo^2 - 2a(∆y)
That equation doesn't make any sense. Vo sin(theta) would give you a velocity, not an angle. Where did you get the 36.44 degrees from?


3. The Attempt at a Solution
I thought Vox = Vx (velocity at the top since Vy = 0) = Vo cos (θ) = 5.0599m/s
That's right. Vox will be the only velocity at the top. You don't say what angle you are using, but if you are using 36.44 degrees, with Vo=16.5 then 5.0599 m/s isn't correct.

Shelly
 
sorry i used
V^2 = (Vosin(θ))^2 - 2g(y-yo) . . .0^2 = (16.5sinθ)^2 - 2*9.8*4.9 and solved for θ = 36.44° . . . so thats where i got the angle which i KNOW is right. . .but how do i get V at the tops which should be Vox
 
458
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I don't know about you, mate, but when I insert 16.5cos(33.44) into my calculator, I don't get 5.06 or 8.29.
 
that is because the angle is 36.44 not 33.44 that is where the 5.06 comes from and the 8.3 comes from :

vy = voy + at
0 = 16.5sin(36.44) - 9.8t and found t = 1.6secs (which could be wrong because 16.5sin(36.44) = -15.7 AND gravity is negative in this case but i just assume that t is going to equal a postive number. . .)
so from there i went ahead and did:

x = x0 + VoxT +.5at^2 but a = o in x direction so the last part is out so. . .
x = xo + VoxT
13.275 = 0 + Vox(1.6) and Vox = 8.3m/s

i got the 13.274 from projectile motion eq.:
D = (Vo^2/9.8)*sin2(θ) = ((16.5^2)/9.8)sin(2*36.44) = 26.44m . . that gives us the total distance y . . .but we want half of that because that is where the max height will be so 26.44/2 = 13.275m
 
458
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Whoops, typo on my part. I meant 36.44. It doesn't give me 5.06 or 8.3 and I don't know why you're trying to use an acceleration in the y-direction to calculate a velocity in the x-direction. (Does velocity in the x-direction change in parabolic motion?)
 
Avodyne
Science Advisor
1,396
85
Vox = Vo*cos(θ) at all times, and at the top of the trajectory Vox is also the speed (since there is no vertical velocity at this point). So once you have the angle, you're essentially done.
 
ya i do have the angle the angle is 36.44 . . .that part is correct . . .and i took 16.5cos36.44 to get 5.0599m/s for Vox. . . isnt that what you get?

and snazzy i was trying to find the time so that is why i used acceleration since i knew everything EXCEPT the time. . solved for time and then used time
 
458
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Convert your calculator from radians to degrees, mate.
 
79
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KMJ - Snazzy is right, recalculate the velocity in the 'y' direction.

16.5(cos 36.44) does not equal 5.06

Snazzy - in projectile motion the 'x' direction velocity remains constant (well, for most of our calculations it does, the problems generally say to ignore wind resistance etc). The 'y' direction changes due to gravity, and is 0 at its highest point.
 
Wow. . . .i knew it had to be something easy like that. . .damn radian button on the . . .forgot to switch it back to degrees. . . so i got 13.2739 . . sorry everything but THANK YOU for sticking with me!!!!!
 
458
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Snazzy - in projectile motion the 'x' direction velocity remains constant (well, for most of our calculations it does, the problems generally say to ignore wind resistance etc). The 'y' direction changes due to gravity, and is 0 at its highest point.
I know, I was asking a question to make the author of the original question think about how he was applying the concepts. It wasn't for you to answer, lol.
 
now how do i show that this problem is solved?
 
50
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sorry i used
V^2 = (Vosin(θ))^2 - 2g(y-yo) . . .0^2 = (16.5sinθ)^2 - 2*9.8*4.9 and solved for θ = 36.44° . . . so thats where i got the angle which i KNOW is right. . .but how do i get V at the tops which should be Vox
Yes, of course, my mistake. That is the angle, sorry about that. The rest of your procedure is correct, except you are getting the wrong answer when you multiply 16.5(cos(36.4)) for some reason.

Shelly
 

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