A ball is thrown towards a vertical wall from a point 2m above the ground and 3m from the wall. the initial velocity of the ball is 20m/s at an angle of 30deg above the horizontal. if the collision of the ball with the ball is perfectly elastic. how far behind the thrower does the ball hit the ground?
The time before hitting the wall is
x =v(ox) •t,
t =x/ v(ox) = x/ v(o) •cosα =
=3/20•0.866 = 0.173 s
v(y) = v(oy) - g•t = v(o) •sin α - g•t = = 20•sin 30o – 9.8•0.173 = 8.3 m/s,
v(x) = v(ox) = v(o)•cos α = 20•0.866 = =17.3 m/s.
v = sqrt(v(x)^2 + v(y)^2) = =sqrt(300+68.9) = 19.2 m/s.
tanβ = v(y)/v(x) = 8.3/17.3 = 0.479,
β =25.6o
v(o) = 19.2 m/s
L = v(o)^2•sin2β/g = 29.34 m.
You have to do this problem in several steps.
First, you need the location and the Velocity direction whre the ball hits the wall.
Then, assume that the horizontal velocity component reverses direction while the vertical velocity component remains the same.
Finally, used the location and direction after the wall bounce as initial conditions to compute where it hits the ground. Relate that to the distance behind the thrower.
I do not have time to go through all the steps for you.
According to Elena's solution, the ball is still travelling upward when it hits the wall. A positive upward angle â is maintained after the bounce. i agree with it up to that point.
The elevation where the ball hits the wall needs to be considered when calculating the range after the bounce. Use of the last L equation is not correct, in my opinion. Also, the question asked for where it hits the ground in relation to the thrower.
Ok.
The answer is 32.53m
And i got that. I didn't use the L equation Instead i found the time taken to reach h=0m then found R using R=(VoCosα)t
THANK YOU ALL SOO MUCHHH!!!!
IT WAS GIVING ME A TOUGH TIME! :)
Therefore the ball will hit the ground at 30.5m behind the thrower.
Ok its corrrct can you do it for me
To find out how far behind the thrower the ball hits the ground, we need to break down the motion of the ball into horizontal and vertical components.
Let's start by finding the time it takes for the ball to hit the ground. We can use the vertical motion equation:
h = ut + (1/2)gt^2
Where:
h = initial height (2m above the ground)
u = initial vertical velocity (vertical component of the initial velocity)
g = acceleration due to gravity (-9.8 m/s^2)
t = time
We can calculate the initial vertical velocity (u) using the given initial velocity and the angle of projection:
u = V * sin(theta)
Where:
V = initial velocity (20m/s)
theta = angle of projection (30 degrees)
Plugging in the values, we get:
u = 20 * sin(30)
u ≈ 10 m/s
Now, let's substitute the values into the equation and solve for time (t):
2 = (10) * t + (1/2) * (-9.8) * t^2
Rearranging the equation, we get a quadratic equation:
4.9t^2 + 10t - 2 = 0
We can solve this equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
t = (-(10) ± √((10)^2 - 4 * 4.9 * (-2))) / (2 * 4.9)
Simplifying the equation, we get:
t ≈ 0.831s or t ≈ -2.031s
Since time cannot be negative in this context, we can discard the negative solution.
Therefore, the time it takes for the ball to hit the ground is approximately 0.831 seconds.
Now, let's find out how far horizontally the ball travels in this time. We can use the horizontal motion equation:
s = ut + (1/2)at^2
Where:
s = horizontal distance traveled
u = initial horizontal velocity (horizontal component of the initial velocity)
a = horizontal acceleration (0, as there is no horizontal force acting on the ball)
t = time
The initial horizontal velocity (u) can be calculated using the given initial velocity and the angle of projection:
u = V * cos(theta)
Where:
V = initial velocity (20m/s)
theta = angle of projection (30 degrees)
Plugging in the values, we get:
u = 20 * cos(30)
u ≈ 17.32 m/s
Now, let's substitute the values into the equation and solve for the horizontal distance (s):
s = (17.32) * t
Plugging in the value of time (t ≈ 0.831s), we get:
s ≈ 14.387m
Therefore, the ball hits the ground approximately 14.387m behind the thrower.
The ball hits the wall when it is moving upwards (before reaching the max height) as the time of the motion to the highest point is t = v(o)sinα/g = 20•0.5/9.8 = 1.02 s. (0.173 s < 1.02 s)
The angle β is measured above horizontal, therefore, at the elastic collision the ball will move at the same angle respectively the horizontal but in opposite direction.
My mistake was that I didn’t pay attention at the words “behind the thrower “ , therefore my answer is 29.34 – 3 = 26.34 m.