Initial velocity u = 40
Angle at launch = 55 degrees
At maximum height v = 0, velocity equation v^2 = u^2 - 2gh,
0 = (40 x sin55)^2 - 2 x 9.81 x h => (40 x 0.819)^2 = 19.62h => h = 32.76^2 /
19.62
Maximum height = 54.7 m
We have v = u - gt => 0 = (40 x sin55) - 9.81 x t => t = 32.76 / 9.81 => t =
3.34 s
Time taken to hit the ground is 2t = 2 x 3.34 = 6.68 s
Distance from castle to trebuchet = utcos55 = 40 x 6.68 x 0.573 = 153.1 m
A single line indicates a bond between 2 atoms .
Answer:
Vf = 210 [m/s]
Av = 105 [m/s]
y = 2205 [m]
Explanation:
To solve this problem we must use the following formula of kinematics.

where:
Vf = final velocity [m/s]
Vo = initial velocity = 0 (released from the rest)
g = gravity acceleration = 10 [m/s²]
t = time = 21 [s]
Vf = 0 + (10*21)
Vf = 210 [m/s]
Note: The positive sign for the gravity acceleration means that the object is falling in the same direction of the gravity acceleration (downwards)
The average speed is defined as the sum of the final speed plus the initial speed divided by two. (the initial velocity is zero)
Av = (210 + 0)/2
Av = 105 [m/s]
To calculate the distance we must use the following equation of kinematics

44100 = 20*y
y = 2205 [m]
Answer:
A. 0N
Explanation:
assume you're pulling in the positive direction, then we have 5 + (-5) = netforce = 0N