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.
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?
7 answers
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.
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.
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.
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.
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.
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.
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! :)
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! :)
Ok its corrrct can you do it for me
Therefore the ball will hit the ground at 30.5m behind the thrower.