A 5.00 kg stone is thrown upward at 7.50 m/s at an angle of 51.0 degrees above the horizontal from the upper edge of a cliff, and it hits the ground 1.50 s later with no air drag. Find the magnitude of its velocity vector just as it reaches the ground.
The horizontal speed does not change. It remains at
7.5 cos51° = 4.72 m/s
The vertical speed is given by v = Vo -9.8t, so it ends up at
7.5 sin51° - 9.8*1.5 = -8.87 m/s
So, the final speed is √(4.72^2 + 8.87^2) = 10.05 m/s
Your school SUBJECT is probably physics.
A car traveling 88 km/h is 110 m behind a truck traveling 75 km/h. how long will it take the car to reach the truck?
Oh, I see we're discussing stones and cliffs. Watch your step, my friend! Now, to find the magnitude of the velocity vector of our stone just as it reaches the ground, we'll need to break it down (no, not literally, that would be messy).
First, let's look at the vertical component of the stone's velocity. We know it was thrown upwards, so initially, it has a positive velocity. The time it takes to reach the ground is 1.50 seconds, so we can use the equation Vf = Vi + at (that's "velocity final equals velocity initial plus acceleration times time") to find it.
Since gravity is pulling the stone downwards, the acceleration is -9.8 m/s^2 (oh, gravity, always trying to ruin our fun!). The initial vertical velocity is 7.50 m/s because it was thrown upwards. So, plugging those values in, we get:
Vf = 7.50 m/s + (-9.8 m/s^2) * 1.50 s
Calculating that gives us the final vertical velocity, which we'll call Vy.
Next, let's find the horizontal component of the velocity. The stone was launched at an angle of 51.0 degrees above the horizontal, so we can use basic trigonometry (oh, hello, old friend trig!) to find the horizontal velocity, which we'll call Vx.
Vx = Vi * cos θ
Plugging in the values, we get:
Vx = 7.50 m/s * cos(51.0 degrees)
Finally, we can use these components to find the magnitude of the velocity vector just before the stone reaches the ground. For that, we'll use the good old Pythagorean theorem!
Velocity magnitude = square root( (Vx)^2 + (Vy)^2 )
Calculating that will give us the answer you seek. But be careful with falling stones, my friend. They can be very "rocky"!
To find the magnitude of the velocity vector when the stone reaches the ground, we need to split the stone's initial velocity into its horizontal and vertical components. Then, we can calculate the time it takes for the stone to reach the ground and find its final velocity using the equations of motion.
Let's start by calculating the initial vertical and horizontal components of the stone's velocity.
The vertical component of the velocity (v_y) can be found using the formula:
v_y = v * sin(θ),
where v is the initial velocity (7.50 m/s) and θ is the angle above the horizontal (51.0 degrees).
v_y = 7.50 m/s * sin(51.0 degrees)
v_y = 7.50 m/s * 0.7771
v_y ≈ 5.83 m/s
The horizontal component of the velocity (v_x) can be found using the formula:
v_x = v * cos(θ),
where v is the initial velocity (7.50 m/s) and θ is the angle above the horizontal (51.0 degrees).
v_x = 7.50 m/s * cos(51.0 degrees)
v_x = 7.50 m/s * 0.6293
v_x ≈ 4.72 m/s
Now, we can calculate the time it takes for the stone to reach the ground. We know that the vertical distance traveled by the stone is the height of the cliff. Let's assume the height of the cliff is h.
Using the equation:
h = v_y * t + (1/2) * g * t^2,
where h is the height of the cliff, v_y is the vertical component of the velocity (5.83 m/s), t is the time it takes for the stone to reach the ground, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the stone is thrown upwards from the upper edge of the cliff, the initial height can be considered zero. So, the equation becomes:
0 = v_y * t + (1/2) * g * t^2.
Rearranging the equation, we get:
(1/2) * g * t^2 = -v_y * t.
Substituting the values, we get:
(1/2) * 9.8 m/s^2 * t^2 = -(5.83 m/s) * t.
Simplifying the equation, we have:
4.9 m/s^2 * t^2 = -5.83 m/s * t.
Dividing both sides of the equation by t, we get:
4.9 m/s^2 * t = -5.83 m/s.
Solving for t, we get:
t = -5.83 m/s / 4.9 m/s^2.
t ≈ -1.19 s.
Since time cannot have a negative value in this context, we neglect the negative sign and take the positive value of t:
t ≈ 1.19 s.
Now that we have found the time it takes for the stone to reach the ground, we can find its final vertical velocity (v_yf) using the equation:
v_yf = v_y + g * t,
where v_yf is the final vertical velocity, v_y is the vertical component of the velocity (5.83 m/s), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes for the stone to reach the ground (1.19 s).
v_yf = 5.83 m/s + (9.8 m/s^2) * (1.19 s)
v_yf ≈ 17.10 m/s.
Finally, we can find the magnitude of the stone's velocity vector just as it reaches the ground:
v_f = √(v_x^2 + v_yf^2),
where v_f is the magnitude of the velocity vector, v_x is the horizontal component of the velocity (4.72 m/s), and v_yf is the final vertical velocity (17.10 m/s).
v_f = √(4.72 m/s)^2 + (17.10 m/s)^2
v_f = √(22.32 m^2/s^2 + 292.41 m^2/s^2)
v_f = √314.73 m^2/s^2
v_f ≈ 17.73 m/s.
Therefore, the magnitude of the velocity vector just as the stone reaches the ground is approximately 17.73 m/s.