A particle is projected with an initial speed of 50 m/s at an angle heta. If the time to reach maximum height is 3s, show that sin heta = 3/5 and find the maximum height and horizontal distance.

At maximum height, the vertical component of velocity ($v_y$) is 0.
The initial vertical velocity is $u_y = u \sin\theta$.
Using the kinematic equation $v_y = u_y - gt_{max}$:

$$0 = u \sin\theta - gt_{max}$$

$$0 = (50)\sin\theta - (10)(3)$$ $$0 = 50\sin\theta - 30$$ $$50\sin\theta = 30$$ $$\sin\theta = \frac{30}{50}$$ $$\sin\theta = \frac{3}{5}$$

This matches the required value.

$$H_{max} = u_y t_{max} - \frac{1}{2} g t_{max}^2$$ $$H_{max} = (30)(3) - \frac{1}{2}(10)(3)^2$$ $$H_{max} = 90 - \frac{1}{2}(10)(9)$$ $$H_{max} = 90 - 45$$ $$H_{max} = \boxed{45 \text{ m}}$$

Since the particle lands at the same horizontal level as it was projected, the total time of flight is twice the time to reach maximum height.

$$T = 2 \times t_{max} = 2 \times 3 = 6 \text{ s}$$

First, find $\cos\theta$ from $\sin\theta = \frac{3}{5}$. Using the identity $\sin^2\theta + \cos^2\theta = 1$:

$$\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$ $$u_x = u \cos\theta = 50 \times \frac{4}{5} = 40 \text{ m/s}$$

$$R = u_x \times T$$ $$R = 40 \times 6$$ $$R = \boxed{240 \text{ m}}$$

The horizontal velocity ($v_x$) remains constant:

$$v_x = u_x = 40 \text{ m/s}$$

The vertical velocity ($v_y$) at $t = 4$ s:

$$v_y = u_y - gt$$ $$v_y = 30 - (10)(4)$$ $$v_y = 30 - 40$$ $$v_y = -10 \text{ m/s}$$

The negative sign indicates the particle is moving downwards.

The direction of motion is given by the angle $\alpha$ that the velocity vector makes with the horizontal.

$$\tan\alpha = \frac{|v_y|}{v_x}$$ $$\tan\alpha = \frac{|-10|}{40} = \frac{10}{40} = \frac{1}{4}$$ $$\alpha = \arctan\left(\frac{1}{4}\right)$$ $$\alpha \approx \boxed{14.04^\circ \text{ below the horizontal}}$$