The length of time in seconds it takes for a ball to return to the ground when tossed vertically upward.
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The length of time in seconds it takes for a ball to return to the ground when tossed vertically upward.
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Answer:
The ball will spend 1/2 the total time going up and the same amount of time coming down.
The ball will spend 1/2 the total time going up and the same amount of time coming down.This is useful to know, because at maximum height the ball has a speed of zero.
The ball will spend 1/2 the total time going up and the same amount of time coming down.This is useful to know, because at maximum height the ball has a speed of zero.Initial speed (u) = 30 m/s end speed (v) = 0
The ball will spend 1/2 the total time going up and the same amount of time coming down.This is useful to know, because at maximum height the ball has a speed of zero.Initial speed (u) = 30 m/s end speed (v) = 0Acceleration (a) = -9.81 m/s^2 time (t) = unknown
The ball will spend 1/2 the total time going up and the same amount of time coming down.This is useful to know, because at maximum height the ball has a speed of zero.Initial speed (u) = 30 m/s end speed (v) = 0Acceleration (a) = -9.81 m/s^2 time (t) = unknownNow we can substitute these values into a well known equation for kinematics, under constant acceleration: v = u + at
The ball will spend 1/2 the total time going up and the same amount of time coming down.This is useful to know, because at maximum height the ball has a speed of zero.Initial speed (u) = 30 m/s end speed (v) = 0Acceleration (a) = -9.81 m/s^2 time (t) = unknownNow we can substitute these values into a well known equation for kinematics, under constant acceleration: v = u + atWe find the time (t) taken to get the top and then double it to get the total time of flight.
The ball will spend 1/2 the total time going up and the same amount of time coming down.This is useful to know, because at maximum height the ball has a speed of zero.Initial speed (u) = 30 m/s end speed (v) = 0Acceleration (a) = -9.81 m/s^2 time (t) = unknownNow we can substitute these values into a well known equation for kinematics, under constant acceleration: v = u + atWe find the time (t) taken to get the top and then double it to get the total time of flight.0 = 30 + (- 9.81*t)
The ball will spend 1/2 the total time going up and the same amount of time coming down.This is useful to know, because at maximum height the ball has a speed of zero.Initial speed (u) = 30 m/s end speed (v) = 0Acceleration (a) = -9.81 m/s^2 time (t) = unknownNow we can substitute these values into a well known equation for kinematics, under constant acceleration: v = u + atWe find the time (t) taken to get the top and then double it to get the total time of flight.0 = 30 + (- 9.81*t)t = 30/9.81
The ball will spend 1/2 the total time going up and the same amount of time coming down.This is useful to know, because at maximum height the ball has a speed of zero.Initial speed (u) = 30 m/s end speed (v) = 0Acceleration (a) = -9.81 m/s^2 time (t) = unknownNow we can substitute these values into a well known equation for kinematics, under constant acceleration: v = u + atWe find the time (t) taken to get the top and then double it to get the total time of flight.0 = 30 + (- 9.81*t)t = 30/9.81t = 3.058 seconds
The ball will spend 1/2 the total time going up and the same amount of time coming down.This is useful to know, because at maximum height the ball has a speed of zero.Initial speed (u) = 30 m/s end speed (v) = 0Acceleration (a) = -9.81 m/s^2 time (t) = unknownNow we can substitute these values into a well known equation for kinematics, under constant acceleration: v = u + atWe find the time (t) taken to get the top and then double it to get the total time of flight.0 = 30 + (- 9.81*t)t = 30/9.81t = 3.058 secondsDouble it for total time in the air = 6.12 seconds
The ball will spend 1/2 the total time going up and the same amount of time coming down.This is useful to know, because at maximum height the ball has a speed of zero.Initial speed (u) = 30 m/s end speed (v) = 0Acceleration (a) = -9.81 m/s^2 time (t) = unknownNow we can substitute these values into a well known equation for kinematics, under constant acceleration: v = u + atWe find the time (t) taken to get the top and then double it to get the total time of flight.0 = 30 + (- 9.81*t)t = 30/9.81t = 3.058 secondsDouble it for total time in the air = 6.12 seconds(2 decimal places)
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