A rock is projected from the edge of the top of
a building with an initial velocity of 16.3 m/s
at an angle of 39
◦
above the horizontal. The
rock strikes the ground a horizontal distance
of 90 m from the base of the building.
The acceleration of gravity is 9.8 m/s
2.
Assume: The ground is level and that the
side of the building is vertical. Neglect air
friction.
What is the horizontal component of the
rock’s velocity when it strikes the ground?
Answer in units of m/s
To find the horizontal component of the rock's velocity when it strikes the ground, we can use basic projectile motion equations.
First, let's analyze the motion of the rock.
We know the following values:
- Initial velocity (v0) = 16.3 m/s
- Angle of projection (θ) = 39 degrees
- Horizontal distance traveled (d) = 90 m
- Acceleration due to gravity (g) = -9.8 m/s^2 (negative because it opposes the motion)
We can break down the initial velocity into its horizontal (vx) and vertical (vy) components using trigonometry. The horizontal component is given by v0 * cos(θ), and the vertical component is given by v0 * sin(θ).
So, vx = 16.3 m/s * cos(39 degrees)
= 16.3 m/s * 0.777
≈ 12.68 m/s
Now, we need to find the time it takes for the rock to travel the horizontal distance of 90 m.
Using the equation d = vx * t, we can rearrange it to find time (t):
t = d / vx
t = 90 m / 12.68 m/s
t ≈ 7.09 s
Finally, we need to find the horizontal component of the velocity at this time (when it strikes the ground). Since there is no horizontal acceleration, the horizontal component of velocity remains constant throughout the motion.
Therefore, the horizontal component of the rock's velocity when it strikes the ground is approximately 12.68 m/s.
To find the horizontal component of the rock's velocity when it strikes the ground, we need to use the initial velocity and the angle of projection.
The horizontal component of velocity can be found using trigonometry. It is given by:
Vx = V_initial * cos(theta)
Where Vx is the horizontal component of velocity, V_initial is the initial velocity, and theta is the angle of projection.
In this case, the initial velocity is 16.3 m/s and the angle of projection is 39 degrees.
Plugging in the values, we have:
Vx = 16.3 m/s * cos(39 degrees)
Using a calculator:
Vx = 16.3 m/s * 0.766
Vx ≈ 12.4818 m/s
Therefore, the horizontal component of the rock’s velocity when it strikes the ground is approximately 12.4818 m/s.