I think D x=vxt because it's equation finding change of x (displacement) and using time
The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.
The formula we use is h = (Vf^2 - Vi^2) / (2*-9.8m/s^2)
Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
The sun emits electromagnetic radiation so I think they are electromagnetic waves.
Answer:
H(max) = (v²/2g)
Explanation:
The maximum height the ball will climb will be when there is no friction at all on the surface of the hill.
Normally, the conservation of kinetic energy (specifically, the work-energy theorem) states that, the change in kinetic energy of a body between two points is equal to the work done in moving the body between the two points.
With no frictional force to do work, all of the initial kinetic emergy is used to climb to the maximum height.
ΔK.E = W
ΔK.E = (final kinetic energy) - (initial kinetic energy)
Final kinetic energy = 0 J, (since the body comes to rest at the height reached)
Initial kinetic energy = (1/2)(m)(v²)
Workdone in moving the body up to the height is done by gravity
W = - mgH
ΔK.E = W
0 - (1/2)(m)(v²) = - mgH
mgH = mv²/2
gH = v²/2
H = v²/2g.
