Depending on the value used for the acceleration due to gravity (9.8 ms-2 or 9.81 ms-2), the ball will go up to 11.48 m or 11.47 m (approx). It will be in the air for a total of 3.06 seconds.
This is a velocity question so u need to use uvaxt
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.
If the person sat on the train their velocity relative to the ground would be 95kph. But he/she is goind 3kph to oppose this. So 95-3 = 92 kph to the north is velocity of person relative to the ground.
The initial velocity is 10 meters/sec and is thrown up against the gravitational pull of the earth. This means that the ball is experiencing a deceleration at the rate of 9.8 meters/sec/sec to bring its final velocity to zero. v^2 - u^2 = 2gs where u is the initial velocity, v the final velocity, g is the acceleration or deceleration, and s is the distance traveled. 0^2 - 10^2 = 2 x (-9.8) x s -100 = -19.6s 100 = 19.6s s = 100/19.6 = 5.102 meters Now v = u + gt where v is the final velocity, u is the initial velocty, g is the acceleration or deceleration, and t is the time. When the ball is thrown up with 10 meters/sec velocity it is acted upon by the deceleration of gravity until its velocity becomes zero. So 0 = 10 - 9.8t or 9.8t = 10 t = 1.020 seconds The time for the ball to go up is 1.020 seconds and the same time is taken for the ball to come back for a total of 2.040 seconds.
20.40
This is a velocity question so u need to use uvaxt
51 seconds.
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
Approx 9.8 metres per second^2, downwards.
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.
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)