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!
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
Assuming the acceleration due to gravity is -9.81 m/s^2, the time it takes for the baseball to hit the ground can be calculated using the formula: time = (final velocity - initial velocity) / acceleration. In this case, the final velocity will be 0 m/s when the baseball hits the ground. Calculating it would give you the time it takes for the baseball to hit the ground.
The speed of the baseball when it hits the ground is 150 m/s, assuming no air resistance. This is because the initial velocity will decrease due to gravity until it reaches zero at the highest point, and then starts to increase again as it falls back down.
This is a velocity question so u need to use uvaxt
To find the initial velocity of the kick, you can use the equation for projectile motion. The maximum height reached by the football is related to the initial vertical velocity component. By using trigonometric functions, you can determine the initial vertical velocity component and then calculate the initial velocity of the kick.
The initial velocity of a projectile affects its range by determining how far the projectile will travel horizontally before hitting the ground. A higher initial velocity will result in a longer range because the projectile has more speed to overcome air resistance and travel further. Conversely, a lower initial velocity will result in a shorter range as the projectile doesn't travel as far before hitting the ground.
The object's initial distance above the ground The object's initial velocity
The velocity of the tomato when it hits the ground will be determined by its initial velocity, the force of gravity acting upon it, and any air resistance. It will likely be accelerating towards the ground due to gravity until it reaches its terminal velocity upon impact.
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.
The speed of the baseball when it hits the ground will be approximately 150 meters per second, neglecting air resistance. The vertical velocity of the ball decreases as it goes up, eventually reaching zero at the peak of its trajectory, then increases as it falls back to the ground.