Answer:
time = 4.4 s
horizontal distance = 488.8 m
Step-by-step explanation:
Height, h = 4 ft
Angle, A = 32 degree above X axis
velocity, u = 131 ft/s
Let the time taken is t.
Use second equation of motion
So, time is 4.4 s.
The horizontal distance is
d = u cos A t
d = 131 cos 32 x 4.4 = 488.8 m
If a department has 11 male employees and 17 female employees, then 198968 different committees of 6 members are possible with the condition that the committee has strictly more female employees than male.
As per the question statement, a department has 11 male employees and 17 female employees.
We re required to calculate the total number different committees that can be formed with 6 members strictly having more female employees than male.
Now for committee of 6 members to have more women than men, there can be two combinations:
(4 women and 2 men)
Or, (5 Women and 1 man).
That is, we will need to calculate the number of combinations we can have by selecting 4 female employees from a group of 17 and 2 Male employees from a group of 11 and the number of combinations we can have by selecting 5 female employees from a group of 17 and 1 Male employee from a group of 11 and add up the two number of combinations to obtain our required answer.
Then comes the most important thing to know to be able to solve this question, i.e., the formula to calculate combinations, which goes as
Therefore, the total number different committees that can be formed with 6 members strictly having more female employees than male is
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Answer:
a) 9.434 m/s
b) i (2+5*t) + (1+8*t-4.905*t²) j
c) t= 8/5 secs
d) 3.598 m/s
e) See explanation
Step-by-step explanation:
Part a)
The speed of the ball can be calculated from the given velocity v = 5i +8j
Taking magnitude of v = 5i + 8j
magnitude (v) = = 9.434 m/s.
Part b)
Using kinematic equation of particle as follows:
Sf = Si + Vi*t + 0.5*a*t² ..... Eq 1
Given: Si = (2i + j) m ; Vi = (5i+8j) m/s; a = -9.81 j m/s²
We evaluate Eq 1:
Sf = (2i+j) + (5i+8j)*t + 0.5*(-9.81j)*t²
We get after combining similar terms:
Sf = i (2+5*t) + (1+8*t-4.905*t²) j ..... Eq 2
Part c)
Using kinematic equation of particle only in i axis as follows we use Eq 1:
Sf = Si + Vi*t + 0.5*a*t²
Given: Si = 2 m ; Sf = 10; Vi = 5 m/s; a = 0;
We evaluate Eq 1:
10 = 2 + 5*t - Solve for t
t = 8/5 seconds
Note: The above is the time t when the ball is due north of (10i+7j) i.e having a position vector of 10 in east direction but unknown in north direction. A point directly above or below 10i + 7j.
Part d)
The interception of ball and the player occurs at the same t = 8/5 secs and @ position vector (10i + aj) where a is a constant needs to be found.
Find a:
Using Eq 2 found in part b:
Sf = i (2+5*t) + (1+8*t-4.905*t²) j
Evaluate @ t= 8/5 secs
Sf = (10) i + (1.2432) j .... Eq 3
To find the speed v of the player when he intercepts the ball at Sf = (10) i + (1.2432) j is evaluated as follows:
v = change in position of player / Time
Hence, v = -3.598 j = 3.598 m/s
Part e)
Friction between the ball and surface from which is launched.