Please solve the above question !
Answers
Given :-
◉ A stone is thrown into air with initial velocity (3i + 4j) m/s.
◉ Acceleration due to gravity, g = 10 m/s²
To Find :-
◉ Horizontal distance covered by the stone before striking the ground.
Solution :-
We have,
⇒ Initial velocity, u = (3i + 4j) m/s
- Horizontal speed = 3 m/s
- Vertical speed = 4 m/s
Now, Let us find the angle between the horizontal and the trajectory of stone.
⇒ Magnitude of Initial velocity, u = √(3² + 4²)
⇒ |u| = √25
⇒ |u| = 5 m/s ...(1)
Let us find the angle between the horizontal and the projectile.
⇒ u cos θ = 3 [ given ]
⇒ cos θ = 3/5 [ from (1) , u = 5 ]
⇒ θ = cos⁻¹ 3/5
⇒ θ = 53°
We know,
⇒ Horizontal Distance covered = u²sin 2θ / g
⇒ h = (5)² (2sinθcosθ) / 2×10 [ given, g = 10 ]
⇒ h = 50sin53°cos53° / 20
⇒ h = 5/2 (4/5 × 3/5)
[ ∵ sin 53° = 4/5 , cos 53° = 3/5 ]
⇒ h = 5/2 × 12/25
⇒ h = 6/5
⇒ h = 1.2 m
So, The horizontal distance covered by the stone is 1.2 m
∴ Option (A) is correct.
option (b) is correct answer...
