The kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s
Given parameters
To find
Kinematics is the part of physics that establishes the relationships between the position, velocity, and acceleration of bodies.
In this case we have a vertical launch
y = y₀ + v₀ t - ½ g t²
Where y and y₀ are the final and initial positions, respectively, v₀ the initial velocity, g the acceleration of gravity (g = 9.8 m / s²) and t the time
With the ball in hand, its position is zero
0 = 0 + v₀ t - ½ g t²
v₀ t - ½ g t² = 0
v₀ = ½ g t
Let's calculate
v₀ = ½ 9.8 2.4
v₀ = 11.76 m / s
In conclusion using kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s
Learn more about vertical launch kinematics here:
brainly.com/question/15068914
[I researched for you, since I am not in that particular level to know that knowledge yet. I assure this is accurate info :)]
The answer is A, red.
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Answer:
W = 0.060 J
v_2 = 0.18 m/s
Explanation:
solution:
for the spring:
W = 1/2*k*x_1^2 - 1/2*k*x_2^2
x_1 = -0.025 m and x_2 = 0
W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2
W = 0.060 J
the work-energy theorem,
W_tot = K_2 - K_1 = ΔK
with K = 1/2*m*v^2
v_2 = √2*W/m
v_2 = 0.18 m/s
Answer:
3.7 m/s
Explanation:
M = 444 kg
U = 5 m/s
m = 344 kg
u = - 5 m/s
Let the velocity of train is V and the car s v after the collision.
As the collision is elastic
By use of conservation of momentum
MU + mu = MV + mv
444 x 5 - 344 x 5 = 444 V + 344 v
500 = 444 V + 344 v
125 = 111 V + 86 v .... (1)
By using the formula of coefficient of restitution ( e = 1 for elastic collision)
-5 - 5 = V - v
V - v = - 10
v = V + 10
Substitute the value of v in equation (1)
125 = 111 V + 86 (V + 10)
125 = 197 V + 860
197 V = - 735
V = - 3.7 m/s
Thus, the speed of first car after collision is 3.7 m/s. negative sign shows that the direction is reverse as before the collision.
Answer:
Explanation:
As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from 
to 
is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.
The total distance covered by the train during the entire journey is the area of the speed-time graph.
Area
As velocity is in 
and time is in 
so the unit of area is 
Hence, the total distance is 
.
