The Baseball will travel upwards until its velocity is zero and then fall back down again.
The acceleration of the baseball is the constant acceleration due to gravity acting in a downwards direction.
The time taken to fall back down will be the same as the time to climb, thus the total time is twice the climb time.
Use Netwon's equations of motion and ignore air resistance, the time to the top of the climb is:
v = u + at
→ t = v - u/a
v = final velocity = 0 m/s at the top of the climb
u = initial velocity = +100 m/s
a = acceleration = -g m/s²
t_top = 0 - (100 m/s) / (-g m/s²) = 100/g s
→ t_total = 2 × t_top = 2 × 100/g s = 200/g s ≈ 200/9.81 s ≈ 20.4 s
since gravity is accelerating down at 9.81 m/s/s, then at the top of its climb velocity is zero and it starts to fall back. velocity = acceleration x t ; 100 = 9.81 x time so time to climb is 10.2 seconds; another 10.2 seconds to fall is a total of 20.4 seconds to hit the ground. That's a long time and I don't think anyone can hit a baseball that hard!
Approx 20.4 seconds.
10
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.
I assume you hit it up from the ground level as well. From Conservation of Energy, it immediately follows that: * If there is no air resistance, when it hits the ground it will, once again, have a speed of 100 meters per second. * Since under usual circumstances there WILL BE air resistance, its speed will be less than 100 meters per second.
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.
20.40
This is a velocity question so u need to use uvaxt
51 seconds.
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.
In a vacuum it would fall back to the same height at the same speed, 150m/s. It would then gain another minuscule fraction of speed as it fell from bat-height down to the ground. In the real world, a falling baseball will reach a maximum speed of around 42 meters per second because the air resistance slows it down. This is called Terminal Velocity.
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)
Take acceleration of gravity = 9.8 m/s2 .The ball stops rising and starts falling in (100/9.8) seconds.It returns to the height of the bat in 2 x (100/9.8) = 20.408 seconds. (rounded)