Question

A projectile is thrown at an angle of 60 degrees with the horizontal with an initial velocity of 147 m/s. Calculate the X and Y component of the velocity of the particle when its inclination becomes 45 degrees.

Answer 1

Explanation:

Here's the answer to your question:

Given,

θ = 60°

u (initial velocity) = 147 m/s.

So, velocity in horizontal direction (along X) = 147 cos 60 (Ux)

Velocity (initial) along vertical direction (along Y) = 147 sin 60 (Uy)

Let after time t , the inclination of the particle with the horizontal is 45°

And, at time t, velocity along X = Vx, and that along Y = Vy.

Now,

$$\begin{lgathered}\frac{V_{x}}{V_{y}} = tan 45 \\\\\implies V_{x} = V_{y}\end{lgathered}$$

Now, since the horizontal component of velocity remains constant,

Vx = Ux = 147 cos 60

And, Vy = Uy - gt (where Vy = Vx = 147 cos 60)

$$\begin{lgathered}\implies 147 cos 60 = 147 sin 60 - 10t \\\\\implies 73.5 = 73.5\sqrt{3} - 10t \\\\\implies 10t = 73.5 (\sqrt{3} - 1) \\\\\implies t = \frac{73.5 (\sqrt{3} - 1)}{10}\end{lgathered}$$

On solving, we get Time t = 5.38 seconds.

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