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How long will it take for a ball to hit the ground if it's thrown horizontally with an initial velocity of 40 meters per second from a height of 10 meters and you assume g is 10 m per second squared?
Wiki User
∙ 16y ago
The horizontal velocity is irrelevant, so ignore it. It doesn't matter whether it's going 4 m/s, 40 m/s or 4000 m/s. The object will hit the ground in the same amount of time. In fact, if you calculate the time it would take for an object to fall 10 meters, you'll have your answer.
y = y0 + v0t - (1/2)gt2
Delta(y) = -(1/2)gt2
t = sqrt(2 x Delta(y)/g)
t = sqrt(2 x 10/ 10)
t = 1.414s
Wiki User
∙ 16y ago
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How do you find the time from acceleration and distance?
You can use the formula for distance covered:distance = (initial velocity) x (time) + (1/2) (acceleration) (time squared) Solve for time. This assumes constant acceleration, by the way. If you assume that the initial velocity is zero, then you can omit the first term on the right. This makes the equation especially easy to solve.
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!
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A stone is thrown horizontally at 8 ms from a cliff 78.4 m high how far from the base of the cliff will the stone strike the ground?
According to the Laws of Physics, Horizontal Velocity is unaffected by Vertical Velocity. Thus you need to first find how long it will take the stone to hit the ground. Assume: Acceleration due to gravity = 9.8 m/s^2 Velocity due to gravity = 9.8x m/s, where x is seconds Displacement due to gravity = 9.8x^2 m, where x is seconds Set Displacement = to 78.4m 78.4m = 9.8x^2 m and solve for.. x = sqrt(8) = 2 sqrt(2) Then to find displacement horizontally, multiply time*velocity horizontally distance from base of cliff = (2 sqrt(2))s * 8 m/s = 16 sqrt(2) m
How long to fall 300 feet?
position final = position initial + (speed initial)(time) + .5(acceleration)(time^2) 0=300 + 0(x) +.5(-9.807)(x^2) 0=300 + 0 + -4.9035(x^2) -300=-4.9035(x^2) -300 / -4.9035 = x^2 61.18 = x^2 x = 7.82 seconds That is to assume that the initial speed is 0. This is incorrect. The initial distance is in feet, 300 feet, but the rate of acceleration is in meters per second squared, (-9.807 meters per second squared). To get the correct answer, the units of measurement must match, either feet or meters. The correct answer is 4.32 seconds.
A baseball was thrown from 1 m hill Assume that the ball had an initial velocity of 70 kmh 19.44 ms at 40 degreesWhat was the initial velocity on x direction?
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A rocket of initial mass m0 is shot vertically upward Assume that the motion occurs under constant gravitational acceleration g Let the initial velocity of the rocket on the surface of the earth be?
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Ignoring air resistance a rock in free fall will have a velocity of what after 4 seconds?
The final velocity is simply acceleration x time. The distance can be calculated from: d = 1/2 acceleration time squared + (initial speed) x time Since the initial speed is assume to be zero, you can simply ignore the second term. The acceleration due to gravity is approximately 9.8 meters per second squared.
What is the relationship between radial force and angular velocity squared?
By radial force, we can assume you mean centripetal force Centripetal force = (Mass)(Radius)(Angular velocity)2
How do you work out height from initial velocity final velocity mass and work done?
If you throw an object up, and assume that air resistance is negligible, knowing the initial velocity is enough. One way to do this is to use conservation of energy. Calculate the energy from the initial velocity, then insert it in the formula for gravitational potential energy.Same for final velocity - the final speed is the same as the initial speed. If you know the work done, you already have the first half of the above steps solved.
How do you find the initial velocity just with the accelaration final velocity and time?
To find acceleration, you take Vi [Initial Velocity] and you subtract if from Vf [Final Velocity.] (Vi - Vf) If they Vi and Vf are already given, you take the two givens and you subtract them from each other. Vi minus Vf. Do not do Vf minus Vi or it will be wrong. After you do that, you divide your answer from T [Time] (Vi - Vf) a= _____ t Once you get your answer, that will be your acceleration.
What is the initial velocity for 25 miles in two minutes?
Anything at all ... BUT if you want to assume a steady speed THEN 12.5 miles per minute = 750 mph
What is the equation used to compute the velocity of a falling object using energy?
Calculate the initial potential energy (PE = mgh). Assume that all of this gets converted to kinetic energy, and solve for velocity (KE = 0.5 mv2).
What is the formula for calculating the distance from the initial speed?
When you say initial speed I assume there will be accelleration. If so you could you: s = ut + 1/2at^2. or s = 1/2(u + v)t where s is distance in meters u is initial velocity in ms v is the final velocity in ms a is accelleration in ms^-2 t is time in s If there is no accelleration then s = ut
How do you find the time from acceleration and distance?
You can use the formula for distance covered:distance = (initial velocity) x (time) + (1/2) (acceleration) (time squared) Solve for time. This assumes constant acceleration, by the way. If you assume that the initial velocity is zero, then you can omit the first term on the right. This makes the equation especially easy to solve.
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!
What is the distance if the acceleration is 0.5 meters per sec square at 20 seconds?
If you assume that the initial speed is zero, you can calculate the distance using the formula:distance = 1/2 x acceleration x time squared
