By the Way, guys, this is based on the equation H= -16t2+vt+s
The answer will depend on what "it" is, and on what its initial velocity is.
No. What counts in this case is the vertical component of the velocity, and the initial vertical velocity is zero, one way or another.
This is a velocity question so u need to use uvaxt
height(s)=1/2(u+v)t 200=1/2(0+v)*1v=400ft/secu = initial speedv=final speed
19.6m
initial velocity of the kick = 28.06 m/s
The object's initial distance above the ground The object's initial velocity
The answer will depend on what "it" is, and on what its initial velocity is.
No. What counts in this case is the vertical component of the velocity, and the initial vertical velocity is zero, one way or another.
To answer this question one would need to know the rock's initial height and velocity.
This result is because the wet ball carries more inertia to weight ratio before hitting the ground , it then compresses, loses some of the liquid weight, becomes lighter, and because of the initial inertial force, can therefore leave the ground at a greater velocity
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
20.40
This is a velocity question so u need to use uvaxt
Ignoring air resistance, the horizontal component of velocity has no connection with, and no effect on, the vertical component. Two bodies that leave the top of the building simultaneously with the same vertical velocity hit the ground at the same time, regardless of their horizontal velocities or their masses. That's the same as saying that a bullet fired horizontally from a gun and a bullet or a stone dropped from the gun's muzzle at the same instant hit the ground at the same instant. Strange but true.
The answer depends on its initial velocity and the height from which its fall to the ground is measured.