The horizontal velocity has no bearing on the time it takes for the ball to fall to the floor and, ignoring the effects of air resistance, will not change throughout the ball's fall, so you know Vx. The vertical velocity right before impact is easily calculated using the standard formula: d - d0 = V0t + [1/2]at2. For this problem, let's assume the floor represents zero height, so the initial height, d0, is 2. Further, substitute -g for a and assume an initial vertical velocity of zero, which changes our equation to 0 - 2 = 0t - [1/2]gt2. Now, solve for t. That gives you the time it takes for the ball to hit the floor. If you divide the distance traveled by that time, you know the average vertical velocity of the ball. Double that, and you have the final vertical velocity! (Do you know why?) Now do the vector addition of the vertical velocity and the horizontal velocity. Remember, the vertical velocity is negative!
To calculate the velocity of the ball just before it hits the ground, we can use the equation of motion: velocity = acceleration x time. The acceleration due to gravity is approximately 9.8 m/s^2. Given the time of 3.0 seconds, we can plug these values into the equation to find the velocity. Therefore, the velocity of the ball just before it hits the ground is 29.4 m/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
yes. If the forces acting on the a moving particle are in equilibrium, (e.g. when a spherical object reaches terminal velocity (neglecting increased air resistance as it gets closer to the ground)) then the particle will be moving at a velocity, that is not 0, yet the velocity will remain constant, and the body will not accelerate or decelerate in any direction, and thus the acceleration is 0.
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.
When a ball is dropped, it starts with an initial velocity of zero. However, as it falls towards the ground, it accelerates due to gravity, causing its velocity to increase. Therefore, the velocity of the ball is non-zero as it falls towards the ground.
Using the acceleration formula, final acceleration is the final velocity minus the initial velocity over elapsed time. Final velocity you gave as 40m/s, and the initial velocity was zero (the apple was stationary on the tree), so the difference is 40 m/s. Divided by the time you gave, 4 s, this will be 10 m/s²
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.
To determine how far a projectile travels horizontally, you need to know the initial velocity of the projectile, the angle at which it is launched, and the acceleration due to gravity. The horizontal range of the projectile can be calculated using the formula: range = (initial velocity squared * sin(2*launch angle)) / acceleration due to gravity.
The acceleration of the ball can be calculated using the formula: acceleration = (final velocity - initial velocity) / time. In this case, the initial velocity is 0 m/s, the final velocity is 20 m/s, and the time is 2 seconds. Therefore, the acceleration would be (20 m/s - 0 m/s) / 2 s = 10 m/s^2.
To calculate the velocity of the ball just before it hits the ground, we can use the equation of motion: velocity = acceleration x time. The acceleration due to gravity is approximately 9.8 m/s^2. Given the time of 3.0 seconds, we can plug these values into the equation to find the velocity. Therefore, the velocity of the ball just before it hits the ground is 29.4 m/s.
The final velocity of the ball when it hits the ground can be calculated using the equation: final velocity = initial velocity + (acceleration due to gravity * time). Assuming the ball was dropped from rest, the initial velocity would be 0 m/s. With the acceleration due to gravity being approximately 9.8 m/s^2, the final velocity would be 32.34 m/s.
Using the kinematic equation ( \text{final velocity}^2 = \text{initial velocity}^2 + 2 \times \text{acceleration} \times \text{distance} ) where final velocity is 0 (at the top) and initial velocity is 2.6 m/s, acceleration due to gravity is -9.8 m/s², and distance is 100 m, you can solve for time to get approximately 5.02 seconds.
Increasing the initial velocity of a projectile will increase both its range and height. Higher initial velocity means the projectile will travel further before hitting the ground, resulting in greater range. Additionally, the increased speed helps the projectile reach a higher peak height before it begins to descend back down.
A parachutist falling before opening the parachute experiences an acceleration due to gravity of approximately 9.81 m/s^2, which is the acceleration due to free fall. This acceleration causes the parachutist's velocity to increase as they fall towards the ground.
To find Chris Bromham's initial velocity when he left the ground, you can use the horizontal distance he traveled, the time he was in the air, and the acceleration due to gravity. The equation to use is: horizontal distance = horizontal velocity * time in the air. By rearranging the equation to solve for the horizontal velocity, you can find Chris Bromham's initial velocity when he left the ground.