Answer:50.39 m/s,
40.46 m
Explanation:
Given
launch angle
Range=240 m
We know that Range 
u=50.39 m/s
(b)maximum height of projectile is given by
H=40.46 m
The solution for this problem would be:(10 - 500x) / (5 - x)
so start by doing:
x(5*50*2) - xV + 5V = 0.02(5*50*2)
500x - xV + 5V = 10
V(5 - x) = 10 - 500x
V = (10 - 500x) / (5 - x)
(V stands for the volume, but leaves us with the expression for x)
First thing to do is to draw the system described above. Then, write an equation for the forces present.
<span>
</span>Σ<span>F = Fg - Ff
</span><span>0 = mgsin</span><span>∅</span><span> - umgcos</span><span>∅</span><span>0 = gsin</span><span>∅</span><span> - ugcos</span><span>∅</span><span>
u = tan</span><span>∅
</span>∅(max) = tan^-1 (u)<span>
</span>
<span> Let’s determine the initial momentum of each car.
#1 = 998 * 20 = 19,960
#2 = 1200 * 17 = 20,400
This is this is total momentum in the x direction before the collision. B is the correct answer. Since momentum is conserved in both directions, this will be total momentum is the x direction after the collision. To prove that this is true, let’s determine the magnitude and direction of the total momentum after the collision.
Since the y axis and the x axis are perpendicular to each other, use the following equation to determine the magnitude of their final momentum.
Final = √(x^2 + y^2) = √(20,400^2 + 19,960^2) = √814,561,600
This is approximately 28,541. To determine the x component, we need to determine the angle of the final momentum. Use the following equation.
Tan θ = y/x = 19,960/20,400 = 499/510
θ = tan^-1 (499/510)
The angle is approximately 43.85˚ counter clockwise from the negative x axis. To determine the x component, multiply the final momentum by the cosine of the angle.
x = √814,561,600 * cos (tan^-1 (499/510) = 20,400</span>
