Answer:
103.5 meters
Explanation:
Given that a stunt person has to jump from a bridge and land on a boat in the water 22.5 m below. The boat is cruising at a constant velocity of 48.3 m/s towards the bridge. The stunt person will jump up at 6.45 m/s as they leave the bridge.
The time the person will jump to a certain spot under the bridge can be calculated by using the formula below:
h = Ut + 1/2gt^2
since the person will fall under gravity, g = 9.8 m/s^2
Also, let assume that the person jump from rest, then, U = 0
Substitute h, U and g into the formula above
22.5 = 1/2 * 9.8 * t^2
22.5 = 4.9t^2
22.5 = 4.9t^2
t^2 = 22.5/4.9
t^2 = 4.59
t = 
t = 2.143 seconds
From definition of speed,
speed = distance /time
Given that the boat is cruising at a constant velocity of 48.3 m/s towards the bridge, substitute the speed and the time to get the distance.
48.3 = distance / 2.143
distance = 48.3 * 2.143
distance = 103.5 m
Therefore, the boat should be 103.5m away from the bridge at the moment the stunt person jumps?
The motion of an airplane when the pilot changes the throttle setting of the engine is described by the first law. The motion of a ball falling down through the atmosphere, or a model rocket being launched up into the atmosphere are both examples of Newton's first law.
I believe it's B. the transmission of heat across matter
Answer:
-0.4 m/s
-3.552 m/s
Explanation:
= Mass of first glider = 0.5 kg
= Mass of second glider = 0.3 kg
= Initial Velocity of first glider = 2 m/s
= Initial Velocity of second glider = -2 m/s
= Final Velocity of first glider
= Final Velocity of second glider = 2 m/s
As the linear momentum of the system is conserved we have

The velocity of glider A is -0.4 m/s
= 0
= -5 m/s
= 0.92 m/s

The velocity of glider A is -3.552 m/s