Arun
Last Activity: 6 Years ago
The wall is ¼ of the range away.
The maximum height is ½ of the range away.
So it takes as long to get from the throw point to the wall as it takes to get from the wall to maximum height.
If the total flight time is T, then the vertical velocity Vv is
Vv at t = 0,
½*Vv at t = T/4,
0 at t = T/2,
-½Vv at t = 3T/4, and
-Vv at t = T.
Then the average velocity during the first interval is 3Vv/4, and the average velocity during the second interval is Vv/4.
Then the vertical distance traveled during the first interval is 3 times the vertical distance during the second interval.
Since the vertical distance traveled during the first interval was 3 m, then the vertical distance traveled during the second interval was 1 m, and the maximum height is 4 m.
range = (V²sin(2Θ))/g → 24m = V²sin(2Θ) / g
max height = (V·sinΘ)² / (2g) → 4m = V²sin²Θ / 2g
Divide range by max height:
24m / 4m = 6 = 2sin(2Θ) / sin²Θ
Use trig identity sin(2Θ) = 2sinΘcosΘ:
6 = 4sinΘcosΘ / sin²Θ = 4cosΘ / sinΘ = 4 / tanΘ
tanΘ = 4/6 = 2/3
Θ = arctan(2/3) = 33.7º
assuming the ball just clears the wall.
