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`        A particle is projected with a speed of 50 metre per second with an angle of 53 degree with the horizontal. find the speed of particle at the instant when its velocity vector makes an angle 53 degree with the vertical and also find that instant`
4 months ago

```							speed at that instant = 50 x 3/5 + 50 x 3/5 = root of 60.uy = 50 x sin53ay = -10vy = 50 x sin37v = u + at30 = 40 – 10tt = 1s
```
3 months ago
```							To determine the angle of the velocity, we use the following equation. Tan θ = Vertical velocity ÷ Horizontal velocity As the particle moves from its initial position to its final position, its vertical velocity decreases at the rate of 9.8 m/s each second. During this time its horizontal velocity is constant. To determine the time when the angle is 45˚, we need to determine the object’s vertical velocity at this angle. Let’s determine the vertical and horizontal components of its initial velocity. Vertical = 50 * sin 60˚, Horizontal = 50 * cos 60˚ = 25 m/s Since horizontal velocity is constant, let’s put 25 m/s and 45˚ into the tangent equation. Tan 45 = Vertical velocity ÷ 25 Vertical velocity = 25 m/s To determine the time for the vertical velocity to decrease from its initial value to 25 m/s, use the following equation. vf = vi – a * t, a = 9.8 25 = 50 * sin 60˚ – 9.8 * t t = (25 – 50 * sin 60˚) ÷ -9.8 This is approximately 1.87 seconds.hope help u!!!
```
3 months ago
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