Answer:
t = 123.59s
Explanation:
For the launch pad section:
Vf = Vo + a*t where Vo=0.
Vf = 35*25 = 875m/s
The distance traveled during the launch:
Now the projectile motion, we know that its initial speed is the speed calculated previously and the initial height is the y-component of the previously calculated distance.
where d= 10937.5m; Vo=875m/s.
Solving for t:
t1 = -11.093s t2 = 98.59s
So, the total time of flight will be:
Assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².
Missing part of the question: determine the magnitude of Porsha's acceleration.
Given the data in the question;
Magnitude of Porsha's acceleration;
To determine the magnitude of Porsha's acceleration, we use Newton's second laws of motion:
Where m is the mass of the object and a is the acceleration.
We substitute the mass of Porsha and the force he used into the equation
Therefore, assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².
Learn more: brainly.com/question/25125444
Answer:
U2 = 47.38m/s = initial velocity of B before impact
Explanation:
An example of the diagram is shown in the attached file because of missing angle of direction in the question
Mass A, B are mass of cars
A = 1965
B =1245
U1 = initial velocity of A = 52km/hr
U2 = initial velocity of B
V = common final velocity of two cars
BU2 = (A + B)*V sin ¤ ...eq1 y plane
AU1 = (A + B) *V cos ¤ ....equ 2plane
From equ 2
V = AU1/(A + B)*cos ¤
Substitute V into equation 1
We have
U2 = (AU1/B)tan ¤ where ¤ = angle of direction which is taken to be 30°
Substitute all parameters to get
U2 = (1965/1245)*52 * tan 30°
U2 = 47.38m/s
