Throwing a ball against a wall for projectile motion

In summary, projectile motion is the curved path that an object takes when it is launched into the air and affected by gravity. When a ball is thrown against a wall, it demonstrates projectile motion by following a curved path back to the ground. Factors such as the angle of the throw, initial velocity, and force of gravity can all affect the projectile motion of the ball. The distance between the wall and the thrower also impacts the time it takes for the ball to return. Finally, horizontal and vertical projectile motion can be observed when throwing a ball against a wall.
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Homework Statement


A woman throws a ball at a vertical wall d = 3.4 m away. The ball is h = 1.7 m above ground when it leaves the woman's hand with an initial velocity of 11 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.)
(a) Where does the ball hit the ground? meters (from the wall)
(b) How long was the ball in the air before it hit the wall? seconds
(c) Where did the ball hit the wall? meters (above the ground)
(d) How long was the ball in the air after it left the wall? seconds
ScreenShot2014-04-26at45348PM.png

Homework Equations



gravity=-9.8m/s2
Time= 0=v0T-½gT2 T<0
t= (v0y+√(v0y2-2gy))/g
distance/range x=x0+V0xt
x/y velocity- v0x=Vcos(θ); v0y=vsin(θ)

The Attempt at a Solution



I did this for a practice problem with different values, and got the correct answers. With these values it says I've gotten all incorrect answers… If someone could please help and verify what I've done wrong!

*I made the person's hand the origin
We're not dealing with forces, and therefor we don't lose velocity due to impact.

V0y= 11sin(45°)=7.778
Solving for total time t: ½(9.81)t2-7.778t-1.7 (h=1.7)
t=+1.324s

(a) I solved for total distance as regular, and subtracted (d=3.4)
v0x=11cos(45)=7.778
X=0+7.778(1.324)-3.4
x=10.29-3.4= 6.898m

(b) i made the x value 3.4 (distance)
3.4=0+7.778t
t=.539s

(c ) I plugged in the time it took to get to the wall to the vertical formula, and added h 1.7
y=7.778(5.39)-½(9.81)(5.39t2)+1.7
y=4.47m

(d)
total time 1.324 - ans(b) = .784s
 
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  • #2
Solving for total time t: ½(9.81)t2-7.778t-1.7 (h=1.7)
Do you mean: $$\frac{1}{2}gT^2-vT\sin\theta-h=0$$ ... it can really help to keep the variables around as long as possible.

Did you realize that ##\sin(45)=\cos(45)=\frac{1}{\sqrt{2}}##?

The idea is to solve:
##x=\frac{1}{\sqrt{2}}vT - d## ... where x is the distance from the wall.

Where T is given by:
$$T=\frac{-v + \sqrt{v^2 +4gh}}{g\sqrt{2}}$$
... it's a + sign in there because we don't want to accept T<0.

So $$x=\frac{-v^2 + v\sqrt{v^2 +4gh}}{2g} -d$$

OK: if this looks like what you did, and you got it wrong, then the only thing I can think of is that you rounded something off somewhere or your calculator threw up a bogus value or something like that.

Check the arithmetic.

Also always possible that the model answer is incorrect.
 
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  • #3
I actually found that I did my arithmetic wrong solving for time (subtracted instead of addition).

Thanks for the help!
 
  • #4
Happens to the best of us :)
 
  • #5


Your solutions for (a) and (c) are incorrect. For (a), you should be solving for the horizontal distance traveled by the ball, not the total distance. The correct equation is x = x0 + v0x * t, where x0 is the initial horizontal position (0 in this case) and t is the time it takes for the ball to reach the wall. Plugging in the values, we get x = 0 + 7.778 * 0.539 = 4.193m.

For (c), you need to use the vertical position formula y = y0 + v0y * t - 1/2 * g * t^2, where y0 is the initial vertical position (1.7m in this case). Plugging in the values, we get y = 1.7 + 7.778 * 0.539 - 1/2 * 9.8 * 0.539^2 = 2.392m.

For (b) and (d), your solutions are correct.
 

Related to Throwing a ball against a wall for projectile motion

1. What is projectile motion?

Projectile motion refers to the motion of an object that is launched into the air and moves along a curved path due to the influence of gravity.

2. How does throwing a ball against a wall demonstrate projectile motion?

When a ball is thrown against a wall, it is launched into the air at an angle and then falls back to the ground, following a curved path due to the force of gravity. This demonstrates the motion of a projectile.

3. What factors affect the projectile motion of a ball thrown against a wall?

The angle at which the ball is thrown, the initial velocity of the ball, and the force of gravity are all factors that can affect the projectile motion of a ball thrown against a wall.

4. How does the distance between the wall and the thrower affect the projectile motion?

The distance between the wall and the thrower affects the time it takes for the ball to travel back to the thrower after hitting the wall. The closer the ball is thrown to the wall, the shorter the travel distance and the quicker the ball will return.

5. What is the difference between horizontal and vertical projectile motion?

Horizontal projectile motion involves the movement of an object in a horizontal direction, while vertical projectile motion involves the movement of an object in a vertical direction. In the case of throwing a ball against a wall, both horizontal and vertical motion are present as the ball moves both horizontally towards the wall and vertically as it falls back to the ground.

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