What is the velocity of a projectile with initial velocity 2m/s at an angle of 30degree to horizontal at the time 0.2s,0.5s,0.8s.Find the maximum height attained by projectile,total horinzontal distance travelled,seconds the projectile remains in air?what is the velocity of touching the ground(hit velocity)of a fall dropped from a height of 3m?

Vo = 2 m/s[30o].

Xo = 2*Cos30 = 1.73 m/s.
Yo = 2*sin30 = 1.0 m/s.

Y = Yo + g*Tr.
Y = 1 -9.8*Tr = 0,
-9.8Tr = -1,
Tr = 0.102 s. = Rise time or time to reach max ht.

Tf = Tr = 0.102 s. = Fall time.

Tr+Tf = 0.102 + 0.102 = 0.204 s. = Time in air.

Y = 1 - 9.8*0.2 = -0.96 m/s.
Y = 1 - 9.8*0.5 = -3.9 m/s.
Y = 1 - 9.8*0.8 = -6.84 m/s.
Negative velocities means the projectile is falling.

Y^2 = Yo^2 + 2g*h. Y = 0, Yo = 1.0 m/s, g = -9.8 m/s^2, h = ?.

Dx = Vo^2*sin(2A)/g. A = 30o, g = 9.8 m/s^2.

V = Vo = 2m/s[30o] = Velocity at which the projectile strikes gnd.

V^2 = Vo^2 + 2g*h. Vo = 0, g = 9.8 m/s^2, h = 3 m, V = ?.

To calculate the velocity of a projectile at a given time, you need to break down the initial velocity into its horizontal and vertical components.

Given:
Initial velocity (v₀) = 2 m/s
Launch angle (θ) = 30°

1. Find the initial vertical velocity (v₀y) using trigonometry:
v₀y = v₀ * sin(θ)

v₀y = 2 * sin(30°)
v₀y = 2 * 0.5
v₀y = 1 m/s

2. Find the initial horizontal velocity (v₀x) using trigonometry:
v₀x = v₀ * cos(θ)

v₀x = 2 * cos(30°)
v₀x = 2 * √3/2
v₀x = √3 m/s

To calculate the velocity of the projectile at a specific time, you use the following equations:

Vertical velocity (v_y) = v₀y - g * t
Horizontal velocity (v_x) = v₀x

Given:
time (t₁) = 0.2 s
time (t₂) = 0.5 s
time (t₃) = 0.8 s

3. Calculate the vertical velocity component at each time:
v_y₁ = v₀y - g * t₁
v_y₂ = v₀y - g * t₂
v_y₃ = v₀y - g * t₃

To find the maximum height attained by the projectile, you need to find the time it takes to reach the maximum height. This occurs when the vertical velocity becomes zero (v_y = 0).

4. Find the time of ascent:
v_y = 0 m/s
0 = v₀y - g * t_ascend

t_ascend = v₀y / g

5. Use the time of ascent to find the maximum height (h):
h = v₀y * t_ascend - (1/2) * g * t_ascend²

6. To find the total horizontal distance traveled (d), you use the horizontal velocity (v₀x) and the time of flight (T). The time of flight is the total time the projectile remains in the air until it hits the ground.

7. Find the time of flight:
T = 2 * t_ascend

8. Find the horizontal distance:
d = v₀x * T

To calculate the velocity of an object falling freely from a height (without any initial velocity) and touching the ground, you can use the laws of motion.

Given:
Height (h) = 3 m

9. Find the final velocity (v) using the equation:
v² = u² + 2 * g * h
Since the object is falling and the initial velocity (u) is zero, the equation simplifies to:
v = √(2 * g * h)

where g is the acceleration due to gravity (approximately 9.8 m/s²).

This should help you calculate the velocity at different times, the maximum height attained, the total horizontal distance traveled by the projectile, and the velocity when the falling object touches the ground.