A firing from the top of the building of height 10m at an angle 30degree to vertical towards ground has the firing velocities of 10m/s.what is the hit velocity?

Cos30 = V/10. V = ?.

To find the hit velocity, we need to break down the initial velocity into its horizontal and vertical components.

Given:
Height of the building (h) = 10m
Angle of firing (θ) = 30 degrees
Initial velocity (v) = 10m/s

First, let's calculate the vertical component of the initial velocity (v_vertical) using the sine function:

v_vertical = v * sin(θ)

v_vertical = 10 * sin(30°)
v_vertical ≈ 5 m/s

Next, let's calculate the time it takes for the object to reach the ground. We can use the vertical motion equation:

h = (1/2) * g * t^2

where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time of flight.

Rearranging the equation, we get:

t^2 = (2 * h) / g
t = √[(2 * 10) / 9.8]
t ≈ √2 ≈ 1.41 s

Now that we have the time of flight, we can calculate the horizontal distance traveled using the horizontal motion equation:

d = v_horizontal * t

Since there is no horizontal acceleration (assuming no air resistance), the horizontal component of velocity (v_horizontal) remains constant throughout the flight.

v_horizontal = v * cos(θ)

v_horizontal = 10 * cos(30°)
v_horizontal ≈ 8.66 m/s

d = 8.66 * 1.41
d ≈ 12.21 m

Finally, we can calculate the hit velocity (v_hit) using Pythagoras' theorem:

v_hit = √(v_horizontal^2 + v_vertical^2)
v_hit = √(8.66^2 + 5^2)
v_hit ≈ √(75.07 + 25)
v_hit ≈ √100.07
v_hit ≈ 10 m/s

Therefore, the hit velocity of the object when it reaches the ground is approximately 10 m/s.