A rock is projected from the edge of the top of a building with an initial velocity of 19.6 m/s at an angle of 31 degrees above the horizontal. The rock strikes the ground a horizontal distance of 77 m from the base of the building. The acceleration of gravity is 9.8 m/s^2.(squared) Assume: The ground is level and that the side of the building is vertical. Neglect air friction. What is the vertical component of the rock's velocity when it strikes the ground? Answer in units of m/s
To find the vertical component of the rock's velocity when it strikes the ground, we can use the equations of motion.
First, we need to find the time it takes for the rock to hit the ground. We can use the horizontal component of velocity and the horizontal distance traveled:
Horizontal component of velocity (Vx) = initial velocity * cos(angle)
Vx = 19.6 m/s * cos(31 degrees) ≈ 16.883 m/s
Now, using the equation of motion for horizontal distance:
Distance = velocity * time
77 m = 16.883 m/s * time
Solving for time:
time = 77 m / 16.883 m/s ≈ 4.56 s
Now, we can use the time to find the vertical component of velocity at that time. The vertical component of velocity changes due to the acceleration of gravity,
Vertical component of velocity (Vy) = initial velocity * sin(angle) - acceleration due to gravity * time
Vy = 19.6 m/s * sin(31 degrees) - 9.8 m/s^2 * 4.56 s
Calculating:
Vy = 10 m/s - 44.688 m/s ≈ -34.688 m/s
The negative sign implies that the velocity is downward. Therefore, the vertical component of the rock's velocity when it strikes the ground is approximately -34.688 m/s.
To find the vertical component of the rock's velocity when it strikes the ground, we can use the equation:
Vf = Vi + at
Where:
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time
First, we need to determine the time it takes for the rock to reach the ground. In order to do that, we will break down the initial velocity into its vertical and horizontal components:
Vi_horizontal = Vi * cos(θ)
Vi_vertical = Vi * sin(θ)
Given:
Vi = 19.6 m/s
θ = 31 degrees
Vi_horizontal = 19.6 * cos(31°)
Vi_horizontal ≈ 16.99 m/s
Vi_vertical = 19.6 * sin(31°)
Vi_vertical ≈ 10.11 m/s
Next, let's find the time of flight (t) by using the horizontal distance traveled by the rock:
d = Vi_horizontal * t
Given:
d = 77 m
Vi_horizontal = 16.99 m/s
77 = 16.99 * t
Solving for t:
t ≈ 4.53 seconds
Now, using the vertical motion equation:
Vf_vertical = Vi_vertical + a * t
Given:
Vi_vertical = 10.11 m/s
a = 9.8 m/s^2
t = 4.53 seconds
Vf_vertical = 10.11 + (9.8 * 4.53)
Vf_vertical ≈ 10.11 + 44.394
Vf_vertical ≈ 54.504 m/s
Therefore, the vertical component of the rock's velocity when it strikes the ground is approximately 54.504 m/s.
Vo = 19.6m/s @ 31deg.
Xo = hor. = 19.6cos31 = 16.80m/s.
Yo = ver. = 19.6sin31 = 10.1m/s.
Dh = Xo*T = Hor. dist.
T = Dh / Xo = 77 / 16.8 = 4.58s = Time
in flight.
t(up) = (Yf - Yo) / g,
t(up) = (0 - 10.1) / -9.8 = 1.03s.
t(dn) = T - t(up) = 4.58 - 1.06=3.55s.
Yf = Yo + gt,
Yf = 0 + 9.8*3.55 = 34.8m/s. = Ver.
component of final velocity.