Answer
correct answer is C
Head-to-tail method
Here head-to-tail method is employed to determine resultant of vectors.
Let horizontal component be 1 and let vertical component be 2
To add these two vectors. Move vector 2 until its tail touches the head of 1. The tail of resultant of these two vectors touch the tail of vector 1 and head of resultant vector will touch the head of vector 2.
Answer:
α = 141.5° (counterclockwise)
Explanation:
If
q₁ = +q
q₂ = -q
q₃ < 0
b = 2*a
We apply Coulomb's Law as follows
F₁₃ = K*q₁*q₃ / d₁₃² = + K*q*q₃ / (2*a)² = + K*q*q₃ / (4*a²)
F₂₃ = K*q₂*q₃ / d₂₃² = - K*q*q₃ / (5*a²)
(d₂₃² = a² + (2a)² = 5*a²)
Then
∅ = tan⁻¹(2a/a) = tan⁻¹(2) = 63.435°
we apply
F₃x = - F₂₃*Cos ∅ = - (K*q*q₃ / (5*a²))* Cos 63.435°
⇒ F₃x = - 0.0894*K*q*q₃ / a²
F₃y = - F₂₃*Sin ∅ + F₁₃
⇒ F₃y = - (K*q*q₃ / (5*a²))* Sin 63.435° + (K*q*q₃ / (4*a²))
⇒ F₃y = 0.0711*K*q*q₃ / a²
Now, we use the formula
α = tan⁻¹(F₃y / F₃x)
⇒ α = tan⁻¹((0.0711*K*q*q₃ / a²) / (- 0.0894*K*q*q₃ / a²)) = - 38.5°
The real angle is
α = 180° - 38.5° = 141.5° (counterclockwise)
initial height of the pole vaulter = 4.2 m
height of the pole vaulter just before it touch the pad = 80 cm
so the total displacement of pole vaulter just before he will touch the pad = 4.2 - 0.80 = 3.4 m
now by kinematics
now after this he will come to rest after compressing the pad by 50 cm
so again we can use kinematics to find its acceleration
so here its acceleration will be - 66.64 m/s^2
