"A woman throws a ball at a vertical wall...." Proj. Motion

In summary, a woman throws a ball at a vertical wall. The ball is h= 3 m above ground when it leaves the woman's hand with an initial velocity of 16 m/s. When the ball hits the wall, the horizontal component of its velocity is reversed; the vertical component remains unchanged. The ball spends 0.53 seconds in the air after it leaves the wall.
  • #1
toboldlygo
32
2
So, I've already answered parts (b) and (c), but I'm struggling with (d). Thanks in advance for any help!

1. Homework Statement

"A woman throws a ball at a vertical wall d = 6.0 m away. The ball is h = 3 m above ground when it leaves the woman's hand with an initial velocity of 16 m/s at 45°. When the ball hits the wall, the horizontal component of its velocity is reversed; the vertical component remains unchanged. (Ignore any effects due to air resistance.)

(d) How long was the ball in the air after it left the wall?"

I've got the height when the ball hits the wall hw= 7.62 m, and the time it took for the ball hit to the wall t = 0.53 sec. I know these are right because they've been graded as such.

Homework Equations



I think the relevant equation is [tex] y = y_0 + v_0t+1/2 g t^2 [/tex].

The Attempt at a Solution


[/B]
I've set y= -7.62, y0=0, and 1/2 g = -4.905; solve for t using the quadratic formula. I'm treating it as if it's a different problem after the ball hits the wall, so it's independent of whatever happens before the ball hits the wall. I don't know if that's right, but it made sense to split the whole problem into two parts: before the ball hits the wall, and after the ball hits the wall.
 
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  • #2
I assume the question is the time it takes to hit the ground after leaving the wall.

I wouldn't say they are independent exactly but you are on the right track with your formula. I would double check your signs and figure out what v0 is.
 
  • #3
paisiello2 said:
I assume the question is the time it takes to hit the ground after leaving the wall.

I wouldn't say they are independent exactly but you are on the right track with your formula. I would double check your signs and figure out what v0 is.

voy would just be sin(45)*16, right? And I double checked the signs (gravity and y are negative according to the coordinate system I'm using) and did the calculation five different times and then also programmed my calculator to do it to make sure, and I keep getting an answer that isn't accepted.
 
  • #4
OK, I guess it might be easier for you to break the problem into two parts. Why don't we call the point where the ball hits the wall "1".

Can you put your formula together at point 1?
 
  • #5
paisiello2 said:
OK, I guess it might be easier for you to break the problem into two parts. Why don't we call the point where the ball hits the wall "1".

Can you put your formula together at point 1?

I'm not quite sure what you mean by that. Do you want to know which formula I used to get the time the ball took to hit the wall? Or do you want me to put in the known quantities into the y equation (which gives me the answer 7.62 m)?
 
  • #6
No, I want you to put together the formula with the right inputs after the ball hits the wall at point 1.
 
  • #7
I just did the problem correctly! I failed to account for gravity when calculating for vy. I was using viy, but I had to use vfy for the "after the ball hits the wall" equation. Thanks for helping me out!
 
  • #8
How can the vertical component remained unchanged in a flight?.
 
  • #9
azizlwl said:
How can the vertical component remained unchanged in a flight?.

It can't, which is what I realized after staring at the problem for an hour and doing it twenty times haha
 

1. How does the ball's velocity affect its motion after being thrown at the wall?

The ball's velocity has a direct impact on its motion after being thrown at the wall. The higher the initial velocity, the farther the ball will travel before returning to the ground.

2. What factors influence the ball's trajectory after being thrown at the wall?

The ball's trajectory is influenced by several factors, including its initial velocity, the angle at which it is thrown, and the force of gravity acting on it. Air resistance and any external forces can also affect the ball's trajectory.

3. How does the ball's angle of projection affect its motion when thrown at the wall?

The angle at which the ball is thrown plays a crucial role in its motion. A lower angle will result in a shorter but higher trajectory, while a higher angle will result in a longer but lower trajectory. The optimal angle for maximum distance is 45 degrees.

4. What happens to the ball's kinetic energy when it reaches its highest point?

The ball's kinetic energy is at its maximum when it is released and decreases as it rises due to the force of gravity. At the highest point of its trajectory, the ball's kinetic energy is converted into potential energy, and then it begins to decrease again as the ball falls back to the ground.

5. How does the height of the wall affect the ball's motion?

The height of the wall will impact the ball's motion in several ways. If the wall is high enough, the ball may not reach it and will continue to travel in a parabolic path until it hits the ground. If the wall is lower, the ball may hit it and bounce off in a different direction. The height of the wall also affects the time it takes for the ball to return to the ground.

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