Projectile Motion of a ball shot into the air

In summary, at a height of 9.1 m, the ball's velocity is observed to be v = (7.9 i + 6.9 j) m/s, with i horizontal and j upward.
  • #1
Kildars
95
0
A ball is shot from the ground into the air. At a height of 9.1 m, the velocity is observed to be v = (7.9 i + 6.9 j) m/s, with i horizontal and j upward.

(a) To what maximum height does the ball rise?
11.5136 m
(b) What total horizontal distance does the ball travel?
m
(c) What is the magnitude of the ball's velocity just before it hits the gound?
m/s
(d) What is the direction of the ball's velocity just before it hits the ground?
° relative to the horizontal

I got A

I don't really know how to start B.. could someone give me a hint?
 
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  • #2
How much time does it spend in the air?
 
  • #3
3.06s?

Using h = 1/2gt^2 and 11.5136 as the H
 
  • #4
OK. Now make use of that to find the horizontal distance traveled.
 
  • #5
\Delta X = Vox*t

Is that the formula I should use?
 
  • #6
That will work.
 
  • #7
What's Vox? If it's 0 it won't work.
 
  • #8
Kildars said:
What's Vox? If it's 0 it won't work.

The velocity in the x-direction remains constant. You only have to read it off from your velocity vector.
 
  • #9
Alright I did \Delta X = 7.9(3.06) and I got 24.174 which is correct.

It's amazing how you guys know this stuff, just like right off the top of your head.
 
  • #10
So for C how would i start doc?
 
  • #11
Kildars said:
So for C how would i start doc?

Write down the equation for the y-component of velocity, that's a good start.
 
  • #12
V(y) = Vosin(theta)-gt

Correcto?
 
  • #13
Kildars said:
V(y) = Vosin(theta)-gt

Correcto?

Correcto. Now, you can find Vo*sin(theta) = Voy directly from y(t) = Voy*t - 1/2*g*t^2, since, at the total time of the flight you calculated, y has to equal zero. When you do that, use the equation for V(y) you wrote down to claculate the y component of the velocity just before the object hits the ground.
 
  • #14
Well I hope I didn't do it wrong, I got Voy = 14.994 .. but I don't know theta?
 
  • #15
Kildars said:
Well I hope I didn't do it wrong, I got Voy = 14.994 .. but I don't know theta?

Read more carefully. We already said that Voy = Vo*sin(theta), so you don't have to know the angle. :smile:
 
  • #16
radou said:
Read more carefully. We already said that Voy = Vo*sin(theta), so you don't have to know the angle. :smile:

V(y) = 14.994 - 9.8(3.06)

Tried it, it doesn't work :(.
 
  • #17
Kildars said:
V(y) = 14.994 - 9.8(3.06)

Tried it, it doesn't work :(.

Of course it doesn't work, since you're asked for the magnitude of the velocity, which means the magnitude of the velocity vector. So, you know Vx already. Now, just plug the result you got on V(y) into the velocity vector and find it's magnitude.
 
  • #18
radou said:
Of course it doesn't work, since you're asked for the magnitude of the velocity, which means the magnitude of the velocity vector. So, you know Vx already. Now, just plug the result you got on V(y) into the velocity vector and find it's magnitude.

Could you try saying that in a different way? I don't understand what that means..

Also check your PMs.
 
  • #19
Kildars said:
Could you try saying that in a different way? I don't understand what that means..

Also check your PMs.

The velocity was given with v = (7.9 i + 6.9 j) = (Vx i + Vy j)m/s, as you wrote in the beginning. Now, as said, since Vx is constant, it equals 7.9 at any moment. Plug in the Vy you calculated (the one you thought was wrong) and you'll get v = (7.9 i + Vy j) m/s. The magnitude of a vector is given with the square root of (7.9^2 + Vy^2).
 
  • #20
radou said:
The velocity was given with v = (7.9 i + 6.9 j) = (Vx i + Vy j)m/s, as you wrote in the beginning. Now, as said, since Vx is constant, it equals 7.9 at any moment. Plug in the Vy you calculated (the one you thought was wrong) and you'll get v = (7.9 i + Vy j) m/s. The magnitude of a vector is given with the square root of (7.9^2 + Vy^2).

Sweet it worked! Thanks!
 

Related to Projectile Motion of a ball shot into the air

1. What is projectile motion?

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

2. What factors affect the projectile motion of a ball?

The factors that affect the projectile motion of a ball include the initial velocity, the angle of projection, the mass of the ball, and the force of gravity.

3. How is the trajectory of a ball in projectile motion calculated?

The trajectory of a ball in projectile motion can be calculated using the equations of motion, which take into account the initial velocity, the angle of projection, and the force of gravity.

4. How does air resistance affect the projectile motion of a ball?

Air resistance can have a significant impact on the trajectory of a ball in projectile motion. It can cause the ball to slow down and deviate from its expected path.

5. What is the maximum height reached by a ball in projectile motion?

The maximum height reached by a ball in projectile motion depends on the initial velocity, the angle of projection, and the force of gravity. It can be calculated using the equation h = (v^2*sin^2θ)/2g, where h is the maximum height, v is the initial velocity, θ is the angle of projection, and g is the force of gravity.

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