# Projectile Motion Problem with thrown ball

• KMjuniormint5
In summary: 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 that's where i got the angle which i KNOW is right. . .but how do i get V at the tops which should be VoxIn summary, the conversation discusses the problem of throwing a ball at a target with a specific height and wanting the ball's velocity to be horizontal upon reaching the target. The equations used include Voy = Vo sin (θ) and V^2 = Vo^2 - 2a

## 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?

## Homework Equations

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

## 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?

KMjuniormint5 said:

## 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.

## 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?

## 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 that's where i got the angle which i KNOW is right. . .but how do i get V at the tops which should be Vox

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

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?)

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. . . isn't 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

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!

chocokat said:
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?

KMjuniormint5 said:
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 that's 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

## 1. What is projectile motion?

Projectile motion is the motion of an object that is thrown, launched, or projected into the air and moves along a curved path under the influence of gravity.

## 2. What factors affect projectile motion?

The factors that affect projectile motion include the initial velocity, launch angle, air resistance, and the force of gravity.

## 3. How do you calculate the maximum height of a projectile?

The maximum height of a projectile can be calculated using the formula h = (v2sin2θ)/2g, where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

## 4. What is the range of a projectile?

The range of a projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the formula R = (v2sin2θ)/g, where v is the initial velocity and θ is the launch angle.

## 5. How does air resistance affect projectile motion?

Air resistance can affect projectile motion by slowing down the object and changing its trajectory. This is because air resistance creates a force that is opposite to the direction of motion, causing the object to lose speed and height as it travels through the air.