A man stands on the roof of a 10.0 -tall building and throws a rock with a velocity of magnitude 24.0 at an angle of 42.0 above the horizontal. You can ignore air resistance. Calculate the magnitude of the velocity of the rock just before it strikes the ground.
The horizontal velocity component will remain
Vx = 24 cos 42 at all times of flight.
They should say what the dimensions are.. I will assume meters/sec fr velocity and meter fr the height h.
The vertical velocity component
which is initially
Vyo = 24 sin 42,
will increase such that
(Vyfinal)^2 = (Vyo)^2 + 2 g h
Solve for Vyf. The magnitude of the final velocity is
Vf = sqrt[Vx^2 + Vyfinal^2]
Here's an easier way to get it: The kinetic energy (1/2) M V^2 increases by the potential energy loss M g h, no matter what angle it is thrown. The mass cancels out.
Therefore
Vf^2/2 = Vo^2/2 + gH
I did it the way that you said: Vx=17.8 Vy=16.1^2 + 2(-9.8)(10m) Vy=63.21 Vf=316.84+3994.24=65.65 but i still got the answer wrong
You got the sign of g wrong. Use +9.8 . Just because gravity goes down does not mean you should use a - sign for g.
Vf^2 = 24^2 + 2*9.8*10
If you insist on using a negative g, then you should use a negative H, because the height decreases. In either case, V^2 must be higher when it hits the ground than it was when it was thrown.
A man stands on the roof of a 10.0 -tall building and throws a rock with a velocity of magnitude 24.0 at an angle of 42.0 above the horizontal. You can ignore air resistance. Calculate the magnitude of the velocity of the rock just before it strikes the ground
What is a 10.0 - tall building?
What units apply to a "magnitude 24.0"?
Is the building 10 stories tall or 10 meters tall or 10 feet tall?
Assuming the building is 10m tall and the velocity is 24m/sec.
The vertical component of the initial velocity is all we need which is Vv = 24(sin42º) = 16.06m/sec.
The time to maximum height above the building derives from Vf = Vo - 9.8t or 0 = 16.06 - 9.8t making t = 1.6338 sec.
The height reached above the building derives from h = Vot - 4.9t^2 or h = 16.06(1.638) - 4.9(1.638)^2 or h = 13.16 meters.
The distance from the maximum height to the ground is d = 10 + 13.16 = 23.16 meters.
The time to reach the ground derives from h = Vot + 4.9t^2 or 23.16 = 0(t) + 4.9t^2 making t = 2.17 sec.
The velocity of impact derives from Vf = Vo + 9.8t or Vf = 0 + 9.8(2.17) = 21.3m/sec.
If, by chance, you meant a 10 story building, its height would more likely be T = H = 10(10)/3.281 = 30.48m.
The distance from the maximum height to the ground is then d = 30.48 + 13.16 = 43.64 meters.
The time to reach the ground derives from h = Vot + 4.9t^2 or 43.64 = 0(t) + 4.9t^2 making t = 2.98 sec.
The velocity of impact derives from Vf = Vo + 9.8t or Vf = 0 + 9.8(2.98) = 29.24m/sec.
I hope this is of some help to you.
To calculate the magnitude of the velocity of the rock just before it strikes the ground, we can use the equations of motion.
First, we need to break down the initial velocity of the rock into its horizontal and vertical components.
Given:
Initial height (h) = 10.0 m
Magnitude of initial velocity (v) = 24.0 m/s
Angle above the horizontal (θ) = 42.0 degrees
The horizontal component of the initial velocity (v_x) can be found using trigonometry:
v_x = v * cos(θ)
The vertical component of the initial velocity (v_y) can also be found using trigonometry:
v_y = v * sin(θ)
Next, we need to find the time it takes for the rock to reach the ground. We can use the equation for vertical motion:
h = v_y * t - 0.5 * g * t^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time it takes for the rock to reach the ground.
Rearranging this equation, we get:
0 = -0.5 * g * t^2 + v_y * t - h
This is a quadratic equation in terms of t. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = -0.5 * g, b = v_y, and c = -h.
Since the rock is thrown upward, we take the positive root of the quadratic equation.
Once we have the time it takes for the rock to reach the ground, we can use it to find the horizontal distance traveled (x) using the equation:
x = v_x * t
Finally, we can calculate the final velocity (v_f) of the rock just before it strikes the ground using the Pythagorean theorem:
v_f = √(v_x^2 + v_y^2)
By plugging in the given values into the equations and following these steps, you can calculate the magnitude of the velocity of the rock just before it strikes the ground.