A projectile is thrown upward at an angle 60^ with the horizontal. The speed of the projectile is 20 m/s when its direct

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A projectile is thrown upward at an angle 60^ with the horizontal. The speed of the projectile is 20 m/s when its direction of motion is 45^ with the horizontal. The initial speed of the projectile is

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Given: The projectile is thrown at an angle $60^\circ$ with the horizontal. At some instant, its speed is $20 \, \text{m/s}$ and its direction of motion is $45^\circ$ with the horizontal.

Find: The initial speed $u$ .

In projectile motion, the horizontal component of velocity remains constant, while the vertical component changes due to gravity.

Resolve the initial velocity into components:

u_x = u\cos60^\circ = \frac{u}{2}

u_y = u\sin60^\circ = \frac{\sqrt{3}u}{2}

When the direction of velocity is $45^\circ$ ,

\frac{v_y}{v_x} = \tan45^\circ = 1

So,

v_y = v_x

Since horizontal velocity remains constant,

v_x = \frac{u}{2}

Hence,

v_y = \frac{u}{2}

Now use the given speed $20 \, \text{m/s}$ :

v = \sqrt{v_x^2 + v_y^2}

20 = \sqrt{\left(\frac{u}{2}\right)^2 + \left(\frac{u}{2}\right)^2}

20 = \frac{u}{\sqrt{2}}

u = 20\sqrt{2}

Therefore, the initial speed is $20\sqrt{2} \, \text{m/s}$ . The correct option is A.

Given: At one instant, the projectile has speed $20 \, \text{m/s}$ and direction $45^\circ$ .

Find: The initial speed $u$ .

At $45^\circ$ , the horizontal and vertical components of velocity are equal. Therefore, if the speed is $20 \, \text{m/s}$ , each component has magnitude:

v_x = v_y = \frac{20}{\sqrt{2}} = 10\sqrt{2}

The horizontal component does not change during projectile motion, so

v_x = u\cos60^\circ = \frac{u}{2}

Thus,

\frac{u}{2} = 10\sqrt{2}

u = 20\sqrt{2}

Therefore, the initial speed is $20\sqrt{2} \, \text{m/s}$ . The correct option is A.

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