Respuesta :
The kinematic relations allow finding the results for the different questions are:
1. The acceleraations is a = 375 m/s²
2. The time is t = 62.5 s
3. The velocity at the ground is v = -14.7 m / s
4. The time to the bottoncliff is 1.7 s
5. The height isy₀ = 122.5 m
6.
A) The velocity is v = 25.2 m / s
B) The height for t=1 s is y = 18.1 m
7.
A) Time in the air is t = 2 s
B) The final veloicty is v = - 19.6 m / s
Kinematics analyzes the movement of bodies, finding relationships between the position, velocity and acceleration of bodies.
Let's look for the answers to a series of questions:
1. They indicate the initial and final velocities of the body and the time to reach it is 20 s, ask how much the acceleration is worth
Let's use the relationship
v = v₀ + a t
a = [tex]\frac{v-v_o}{t}[/tex]
a = [tex]\frac{17500-10000}{20}[/tex]
a = 375 m/s²
2. What is the time it takes to travel a distance of 125 m with an acceleration of 4 m / s²
x = v₀ t + ½ a t²
As it leaves the direction its initial velocity is zero
x = ½ to t²
t = [tex]\sqrt{\frac{2x}{a} }[/tex]
t = [tex]\sqrt{\frac{2 \ 125}{4} }[/tex]
t = 62.5 s
3. A pencil falls from a height and reaches the floor in a time of 1.5 s what is the speed when reaching the floor
v = v₀ - g t
As it is released its initial velocity is zero
v = -g t
v = - 9.8 1.5
v = -14.7 m / s
The negative sign indicates that the velocity is directed downwards.
4. From the edge of a cliff 40 m high, an apple is thrown downward with an initial velocity of 15 m / s. How long does it take to get to the bottom?
y = y₀ + v₀ t - ½ g t²
In this case the initial velocity is negative because it is directed downwards and when it reaches the floor its height is zero, let us substitute
Suppose the acceleration is g = 10 m / s²
0 = 40 - 15 t - ½ 10 t²
0 = 40 - 15 t - 5 t²
let's solve the quadratic equation
t² + 3 t - 8 = 0
t = -3 + Ts 9 + 4 8/2
t = -3 + 6. 4/2
t₁ = - 4.9 s
t₂ = 1.7 s
The time must be a positive quantity, so the correct answer is 1.7 s
5. A stone is dropped from the ceiling and it takes a time 5 s to reach the floor, which is the height of the ceiling
y = y₀ +v₀ t - ½ g t²
as it is released from the ceiling its initial velocity is zero and the height upon reaching the floor is zero
0 = y₀ + 0 - ½ g t²
y₀ = ½ g t²
y₀ = ½ 9.8 5²
y₀ = 122.5 m
6. A man at a height of 30 m throws a ball downward at 7 m / s
A) The speed when reaching the ground
B) The height of the ball when a time of 1 s has fallen.
A) Let's use the kinematics relation
v² = v₀² - 2 g (y -y₀)
v = 7² - 2 9.8 (0 -30)
v = 25.2 m / s
B) y = y₀ + v₀ t - ½ g t²
y = 30 7 1 - ½ 9.8 1²
y = 18.1 m
7. A watermelon is dropped from rest from the top of a 20m cliff.
A) time in the air
B) ground speed
A) Let's use the relation
y = y₀ + v₀ t - ½ g t²
It is released its initial velocity is zero and when it reaches the ground its height is zero (y = 0)
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{\frac{2y_o}{g} }[/tex]
t = [tex]\sqrt{\frac{2 \ 20}{9.8} }[/tex]
t = 2 s
B) the speed when reaching the ground
v = v₀ - g t
v = 0 - 9.8 2
v = - 19.6 m / s
The negative sign indicates that the speed is down
In conclusion using the kinematic relations we can find the results for the different questions are:
1. The acceleraations is a = 375 m/s²
2. The time is t = 62.5 s
3. The velocity at the ground is v = -14.7 m / s
4. The time to the bottoncliff is 1.7 s
5. The height isy₀ = 122.5 m
6.
A) The velocity is v = 25.2 m / s
B) The height for t=1 s is y = 18.1 m
7)
A) Time in the air is t = 2 s
B) Te final veloicty is v = - 19.6 m / s
Learn more here: brainly.com/question/15068914
