A) The longest horizontal distance is reached at 45 degrees angle. This is true for any projectile launch.
B) First, calculate fligth time (using the vertical motion) and then calculate the horizontal movement.
Flight time = 2* ascendent time
ascendent time => final vertical velocity, Vy, = 0
sin(45) = Voy / Vo => Voy = Vosin(45) = 25.5 m/s * (√2) / 2 = 18.03 m/s
Vy = Voy - gt = 0 => Voy = gt = t = Voy / g
Use g = 10 m/s^ (it is an aproximation, because the actual value is about 9.81 m/s^2 depending on the latitud)
t = 18.03 m/s / 10 m/s^2 = 1.83 s
This is the ascendant time going upward.
The flight time is 2*1.83 = 3.66 s
Horizontal motion
Horizontal velocity = Vx = constant = Vox = Vo*cos(45) = 18.03 m/s
Vx = x / t => x = Vx*t
Horizontal distance = xmax = 18.03m/s*3.66 s = 65.99 m
c) The time the ballon will be in the air was calculated in the part B, it is 18.03 s
Answer: 3=24/8 24/8 - 1/8 = 23/8 or 2 7/8
Step-by-step explanation:
Exchange 3 into 24/8 which is still 3 but in 8th’s. Then subtract 1/8 from 24/8 which is 23/8 or 2 7/8
Answer:
B. a pair of alternate exterior angles with measures of 130° and 50°
Step-by-step explanation:
Alternate exterior angles formed when two parallel lines are cut across by the same transversal line, are said to be congruent.
Therefore, a pair of alternate exterior angles cannot have different angle measures of 130° and 50°. Rather, both alternate exterior angles should be of equal measure of degree.
All other scenarios are possible result except option B: "a pair of alternate exterior angles with measures of 130° and 50°".
