Respuesta :
Answer:
a.)How far from the building does the ball hit the ground?
72 m
b.)How long does it take to hit the ground?
6 s
c.)What is its speed when it hits the ground?
63.48 m/s
Explanation:
When an object is in free fall with no air resistance present or wind acting on it it only is under the influence of gravity. Gravity acts straight down and accelerates the object all the way to the ground. If there is air resistance or a wind force on the object it will have an acceleration that isn't 9.8 m/s^2
The kinematics allows to find the results for the questions are:
- The time is t = 6 s
- The horizontal distance is x = 72 m
- The velocity at the gound is v = 63.5 m / s with direction θ= 292.2º
Kinematics studies the movement of bodies, finding relationships between their position, speed and acceleration.
Newton's second law states that force is proportional to the product of mass and acceleration of the body.
F = ma
Wher F is the force, m is the mass and a is the acceleration
They indicate that there is a horizontal force of 12 N on the body, we look for the acceleration of the horizontal force ₓ
[tex]a_x = \frac{F_x}{m}\\\\a_x = \frac{12}{3}[/tex]
aₓ = 4.00 m / s²
In this case we can see that we have a motion in two dimensions with acceleration in each direction, on the vertical axis that is positive upwards the gravity acceleration acts downwards and on the horizontal axis the acceleration is 4.00 m / s²
a) Let's find the time it takes to reach the ground
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
When you hit the ground the height is zero and its initial height is 176.4 m, in addition to releasing the body, the initial vertical velocity is zero
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{\frac{2 y_o }{ g} }[/tex]
t = [tex]\sqrt{\frac{2 \ 176.4 }{ 9.8} }[/tex]
t = 6 s
b) How far it hits the building
x = x₀ + v₀ₓ t + ½ aₓ t²
Let's locate it is zero in the building therefore its initial position is zero and as the body is released its initial horizontal velocity is zero
x = 0 + 0 + ½ aₓ t²
x = ½ 4.00 6²
x = 72 m
c) The speed when reaching the ground
We look for the vertical speed
[tex]v_y = v_{oy} - gt[/tex]
[tex]v_y[/tex] = 0 - 9.8 6
[tex]v_y[/tex] = -58.8 m / s
We look for horizontal speed
vₓ = v₀ₓ + aₓ t
vₓ = 0 + 4.00 6
vₓ = 24 m / s
Let's use the Pythagoras' theorem to find the modulus of the velocity
[tex]v= \sqrt{v_x^2 + v_y^2}[/tex]
v = [tex]\sqrt{24^2 + 58.8^2 }[/tex]
v = 63.5 m / s
We use trigonometry to find the direction
tan θ' = [tex]\frac{v_y}{v_x}[/tex]
θ' = tan⁻¹ [tex]\frac{v_y}{v_x}[/tex]
θ' = tan⁻¹ ( [tex]\frac{-58.8}{24}[/tex] )
θ' = -37.8º
If we want to measure this angle from the positive side of the x axis in a counterclockwise direction
θ = 360 - θ'
θ = 360 - 37.8
θ = 292.2º
In conclusion using kinematics we can find the results for the questions are:
- The time is t = 6 s
- The horizontal distance is x = 72 m
- The velocity is v = 63.5 m / s with direction θ = 292.2º
Learn more here: brainly.com/question/13518805
