Time before projectile hits wall
= 88.2 m / 29.4 m/s = 3 seconds
Vertical velocity of projectile after three seconds
= 3*9.8 = 29.4 m/s
Horizontal velocity of projectile after three seconds, assuming no air resistance
= 29.4 m/s (given)
Conclusion:
velocity of projectile when it hits the wall
= < 29.4, -29.4> m/s
= sqrt(29.4^2+29.4^2) m/s east-bound at 45 degrees below horizontal
= 41.58 m/s east-bound at 45 degrees below horizontal.
Answer:
Terminal velocity of object = 12.58 m/s
Explanation:
We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.
Gravitational force = mg = 80 * 9.8 = 784 N
Drag force = 
Equating both, we have

So v = 12.58 m/s or v = -15.58 m/s ( not possible)
So terminal velocity of object = 12.58 m/s
Answer:
7.9 
Explanation:
Take the fact that mass is inversely proportional to accelertation:
m ∝ a
Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

Using algebra, we can rearrange our equation to find the final acceleration,
:

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:
0.85g * 9.8
= 8.33
Plug everything in:
7.9
= 
(1590kg the initial weight plus the weight of the added passenger)
Answer:
E) True. Ball B will go four times as high as ball A because it had four times the initial kinetic energ
Explanation:
To answer the final statements, let's pose the solution of the exercise
Energy is conserved
Initial
Em₀ = K
Em₀ = ½ m v²
Final
Emf = U = mg h
Em₀ = emf
½ m v² = mgh
h = v² / 2g
For ball A
h_A = v² / 2g
For ball B
h_B = (2v)² / 2g
h_B = 4 (v² / 2g) = 4 h_A
Let's review the claims
A) False. The neck acceleration is zero, it has the value of the acceleration of gravity
B) False. Ball B goes higher
C) False has 4 times the gravitational potential energy than ball A
D) False. It goes 4 times higher
E) True.
B. 1520 is the difference between their weights.