Question

(g = 10 m/sec)
A projectile is thrown at an angle 30° from the horizontal ground with velocity 10 m/s. Find angle
between displacement vector and velocity vector at time t = 1 sec.​

Answer 1

Given:

A projectile is thrown at an angle x = 30° from the horizontal.

Initial velocity, U = 10m/s

To Find:

the angle  between displacement vector and velocity vector at time t = 1 sec.​

Solution:

Forces acting on the projectile is gravity downwards.

Therefore displacement vector after t sec ,

• s  = Uxt + 1/2axt² i + Uyt + 1/2ayt² j
• s = Ucos x t + 0 i + Usin x t - 0.5xgt²
• S = 10cos(30)t i + 10sin(30)t - 5t²
• S = 5√3 t i + 5t - 5t² j

Velocity vector after t sec,

• V = Ux + axt i + Uy + ayt j
• V = 10cos(30)i + 10sin(30) -gt j
• V = 5√3 i + 5 - 10t j

At t = 1 sec

Displacement vector , S = 5√3i + 0j

Velocity Vector, V = 5√3i -5j

Let angle between both these vectors be A.

Then, dot product is given by,

• S.V = |S||V|cosA
• (5√3)² - 0 = |5√3|√($5\sqrt{3} ^{2} + 5^{2}$)cosA
• 5√3 /10 = cos A
• cos A = √3/2
• A = $cos^{-1}\frac{\sqrt{3} }{2}$ = 30 degree