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# A ball is thrown from the ground to clear a wall 3m high at a distance of 6m and falls 18m away from the wall. Find the angle of projection of ball

Grade:12

## 3 Answers

Arun
25763 Points
2 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.

naman
30 Points
2 years ago
Use the equation of trajectory for solvingvthe question:-
Taking y as 3 and x as 6 and then find the value of theta
Use the formula y=x tan theta(1-X/R)...
Prateek srivastava
13 Points
2 years ago
Use eqn of trajectory y = x(1-x/R)tan# taking x = 6 and y = 3 and r as 24 and find the value of # it would come out as arctan(2/3) which is the required angle of projection.

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