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.
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,
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 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 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! :)
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