Question

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

Answer 1

Answer:

tan thita =2/3

Explanation:

bcoz,

from equation trajectory =x tan =(1-x/R)

Therefore, 6 tantheta (1-1/4)=tantheta=2/3.

I hope this is helpful for you

Answer 2

Given :-

Vertical distance = y = 3 m.

Distance to be travelled = x = 6 m.

But actual,

Distance travelled = 18 m

Total distance = 6 + 18 = 24 m.

R = u² Sin 2Ø/g = 24 m

2u² SinØ.CosØ/g = 24

u² SinØ.CosØ/g = 12

g/u² = SinØ.CosØ/12 -----(1)

Now,

Using Trajectory Equation.

y = x tanØ - gx²/2u² Cos²Ø

3 = 6 tanØ -36g/2u² Cos²Ø ---(2)

Putting eq (1) in eq (2).

3 = 6 tanØ - 36 SinØ.CosØ/12 × 2 Cos²Ø

3 = 6 tanØ - 3/2 tanØ

6 = 12 tanØ - 3 tanØ

9 tanØ = 6

tanØ = 6/9

Ø = tan-¹ (2/3)

Hence,

The angle of Projection = Ø = tan-¹(2/3).