Question

a projectile is thrown in the upward direction making an angle of 60 with horizontal direction with a velocity of 147 m/s . then the time after which its inclination with the horizontal is 45 degree, is?

Answer 1

HELLO THERE!

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,

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

Now, since the horizontal component of velocity remains constant,

Vx = Ux = 147 cos 60

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

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

On solving, we get Time t = 5.38 seconds.

THANKS!

Answer 2

Answer:

Hope it helps you

Thanks You