Question

In an equilateral triangle PQR, PX is the altitude on QR. Prove that: 3PQ^2 = 4PX^2

Answer 1

We'll apply the 2nd Equation of Motion to find Acceleration.

2nd Equation of Motion → \sf s = \dfrac{1}{2}at^{2}+uts=

2

1

at

2

+ut

Substitute the values and find the value of 'a'

⇒ 80 = \dfrac{1}{2}a(5)^{2} + 0(5)80=

2

1

a(5)

2

+0(5)

⇒ 80 = \dfrac{1}{2}a(25) + 080=

2

1

a(25)+0

⇒ 80 = 12.5a80=12.5a

⇒ a = \dfrac{80}{12.5}a=

12.5

80

⇒ a = 6.4 \ m/s^{2}a=6.4 m/s

2

∴ The acceleration would be 6.4 m/s²