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The height h of a ball thrown into the air with an initial vertical velocity of 48 feet per second from a height of 5 feet above the ground is given by the equation?
Wiki User
∙ 13y ago
At any time 't' seconds after the ball is released,
until it hits the ground,
h = 5 + 48 t - 16.1 t2
Wiki User
∙ 13y ago
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Will a ball drop rest reach the ground quicker than the one lunched from the same height but with and initial horizontal 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.
If a cricket jumps off the ground with an initial vertical velocity of 4 ft per second what is an equation the height in ft of the crisket as a function of time in seconds since it jumped?
4ft*Ns=H
If a ball rolls off the edge of a table two meters above the floor and with an initial velocity of 20 meters per second what is the ball's acceleration and velocity just before it hits the ground?
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!
Question-1 A body is thrown vertically upwards The distance st above the ground after t seconds is given by st 20t-5t2 meters?
The equation for vertical motion is y = v0t + .5at2. y is vertical displacement v0 is initial vertical velocity a is acceleration (in meters, normal gravitational acceleration is about -9.8 m/s/s, assuming positive y is upward displacement and negative y is downward displacement)
How long does it take to hit the ground from 28000 feet high?
The answer will depend on what "it" is, and on what its initial velocity is.
Will a ball drop rest reach the ground quicker than the one lunched from the same height but with and initial horizontal 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.
If a cricket jumps off the ground with an initial vertical velocity of 4 ft per second what is an equation the height in ft of the crisket as a function of time in seconds since it jumped?
4ft*Ns=H
If a ball rolls off the edge of a table two meters above the floor and with an initial velocity of 20 meters per second what is the ball's acceleration and velocity just before it hits the ground?
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!
A projectile is fired horizontally from a gun that is 71.0 m above flat ground emerging from the gun with a speed of 250 ms What is the magnitude of the vertical component of its velocity as it st?
If it's fired horizontally, then its initial vertical velocity is zero. After that, the vertical velocityincreases by 9.8 meters per second every second, directed downward, and the projectile hitsthe ground after roughly 3.8 seconds.Exactly the same vertical motion as if it were dropped from the gun muzzle, with no horizontal velocity.
A base jumper jumps off of a cliff that is 200 meters high How long will is take before he hits the ground?
you need a quadratic equation for this ½ at2 + vot - s = 0 vertical acceleration (a) is gravity (-9.8ms-2) initial vertical velocity is 0 his vertical height above ground is 200 (s=200) pop all that in the equation and you're done yep... and I'm sorry but I've had to delete my quadratic formula off my calculator and I've finished maths for the year and can't be stuffed doing it by hand.. you know the quadratic formula.. have fun :)
How do you find time of flight from initial velocity and an angle at which an object is thrown?
If you're willing to ignore the effect of air resistance, then the answer is as follows: The object's horizontal velocity remains constant (at least until it eventually hits the ground). The vertical component of the object's initial velocity ... call it V(i) ... is the (total initial velocity) multipled by the (sine of the initial angle above the horizontal). Beginning at the time of the toss, the magnitude of the vertical component of velocity is V = V(i) - 1/2gT2. T = number of seconds after the toss g = acceleration of gravity = approx 32 ft/sec2 or 9.8 m/sec2
Question-1 A body is thrown vertically upwards The distance st above the ground after t seconds is given by st 20t-5t2 meters?
The equation for vertical motion is y = v0t + .5at2. y is vertical displacement v0 is initial vertical velocity a is acceleration (in meters, normal gravitational acceleration is about -9.8 m/s/s, assuming positive y is upward displacement and negative y is downward displacement)
A guy drove a car from a flat ground on top of a building and laded on the next one that's 100m away several floors down. If the car landed with an angle 30 degrees to x axis, find initial horizontal speed of car?
Initial horizontal velocity = ux (Note this velocity does not affect by gravity so stays the same throughout the jump) Initial vertical velocity = uy = 0 time to travel horizontal distance, t = (horizontal distance)/(horizontal velocity) t = 100/ux This is the same time the car takes to travel vertical distance. Using one of the equation of motion vertical landing velocity, vy = uy + gt vy = 0 + (9.81)(100/ux) vy = 981/ux angle of landing = 30° tan30° = vy/ux tan30° = (981/ux)/ux (ux)² = 981/tan30° ux = 41.22 m/s
What is A football kicked off at an angle 20 from the ground reaches a maximum height of 4.7 m What is the initial velocity of the kick?
initial velocity of the kick = 28.06 m/s
Which information is needed to find the time a projectile is in motion?
The object's initial distance above the ground The object's initial velocity
How long does it take to hit the ground from 28000 feet high?
The answer will depend on what "it" is, and on what its initial velocity is.
When a body is projected simultaneously from the top of the building with different initial horizontal velocity which of the two will hit the ground first?
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.
