An example you could use for this would be if you threw a pencil across the room 10m away. the pencil would have travelled 10m(N). you could determine the objects speed by using the formula speed=distance/time. you would have had to time how long it took for the pencil to hit the 10m marking and divide 10 by however many seconds it took.
Formula: PE = mgh
m = 20 kg, g = 9.8 m/s^2 h= ?
1000 = 20 * 9.8 * h
1000 = 196h
h = 5.10204082
The height is around 5m
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
Answer:
Bow Line
Explanation:
If the wind or current is pushing your boat away from the dock, bow line should be secured first.
1- We should cast off the bow and stern lines.
2-With the help of an oar or boat hook, keep the boat clear of the dock.
3-Leave the boat on its own for sometime and let the wind or current carry the boat away from the dock.
4 - As you see there is sufficient clearance, shift into forward gear and slowly leave the area.
KE= 1/2MV^2 - equation
KE= 1/2 (2 kg)(15 m/s)^2 - plug it to the equation
KE= (1 kg)(225 m/s) - multiply
KE= 225 J - Answer (letter D)
Hope this helps
