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Grade 11Mechanics

A ball is thrown from ground to just clear a well 3m hight at a distance 6m and ball falls 18 m away from the wall find angle of projection

Harsh goswami

8 Years agoGrade 11

1 Answer

Arun

8 Years ago

Dear student

he 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

Regards

Arun (askIITians forum expert)

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