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When throw an object upward and return after 6 seconds what was its initial velocity?
Use the equation vf = vi + gt, where vf is final velocity, vi is initial velocity, g is acceleration due to gravity, and t is time.
Known:
vf = 0m/s
g = -9.8m/s2
t = 6s
Unknown:
vi
Equation:
vf = vi + gt
Solution:
vi = vf - gt
vi = 0m/s - (-9.8m/s2)(6s) = 58.8m/s = 60m/s (rounded to 1 significant figure)
What else can I help you with?
When will a stone return to the groun if it is thrown upward with an initial velocity of 16ms?
The stone will return to the ground when its vertical velocity becomes zero and it starts to fall back down. The time it takes for this to happen can be calculated using kinematic equations. In this case, the time can be found by setting the vertical velocity to zero and solving for time.
To celebrate a victory a pitcher throws her glove straight upward with an initial speed of 6.2 How long does it take for the glove to return to the pitcher?
I am assuming the initial speed is 6.2 m/s Let upward motion be positive! Gravity decreases the speed by 9.8 m/s each second Acceleration due to gravity = -9.8 m/s each second (negative because gravity accelerates objects downward) Find time to reach the top of the path! Final velocity at the top = 0 m/s Initial velocity = 6.2 m/s Final velocity = Initial velocity + acceleration * time Time - = (final velocity - initial velocity) ÷ acceleration Time = (0 - 6.2) ÷ -9.8 = 0.633 seconds (to reach top) The path is symmetrical. 0.633 seconds to reach top and 0.633 seconds to reach glove again. Total time = 12.66 seconds
Find speed throwing object straight up and return 6 seconds later?
To find the speed at which the object was thrown upward, we need to know the acceleration due to gravity. Assuming g ≈ 9.81 m/s², the initial speed of the object can be calculated using the equation v = u - gt, where v is the final speed (0 m/s when it returns), u is the initial speed, g is the acceleration due to gravity, and t is the time (6 seconds). This calculation will give you the initial speed at which the object was thrown upward.
When air resistance is not negigble how does return speed compare with the initial speed?
When air resistance is not negligible, the return speed of an object will be slower than the initial speed because air resistance acts in the opposite direction of the object's motion, slowing it down. This results in a decrease in the object's speed over time.
When throwing a ball straight up when in the air with an initial velocity of 10 meters per second what high will it go and how long will it take to return to the ground?
The initial velocity is 10 meters/sec and is thrown up against the gravitational pull of the earth. This means that the ball is experiencing a deceleration at the rate of 9.8 meters/sec/sec to bring its final velocity to zero. v^2 - u^2 = 2gs where u is the initial velocity, v the final velocity, g is the acceleration or deceleration, and s is the distance traveled. 0^2 - 10^2 = 2 x (-9.8) x s -100 = -19.6s 100 = 19.6s s = 100/19.6 = 5.102 meters Now v = u + gt where v is the final velocity, u is the initial velocty, g is the acceleration or deceleration, and t is the time. When the ball is thrown up with 10 meters/sec velocity it is acted upon by the deceleration of gravity until its velocity becomes zero. So 0 = 10 - 9.8t or 9.8t = 10 t = 1.020 seconds The time for the ball to go up is 1.020 seconds and the same time is taken for the ball to come back for a total of 2.040 seconds.
What is the formula used to calculate the velocity of an object?
The average velocity in a particular direction = distance travelled in that direction / time taken. Velocity is a vector so the direction is important. If I go from A to B and then return to A my average velocity will be zero. My speed, on the other hand, will not be zero.
An object is thrown upward will return to its initial position with same speed if air resistance is negligeble what heppened if air resistance is appreciable?
it would be slower
A man standing at a high tower of 60m and throws the ball upward with a velocity of 20m per second.Find after what time it will pass the man?
If we can use the approximation that we're at or near the surface of the earth and gravity = 9.8 m/s2, then we don't really care where the man is standing when he initiates this experiment ... it doesn't matter.All we're really trying to determine is how long it takes a ball tossed upward at 20m/s to return to its original altitude.And that's just double the time it takes a dropped object to reach a downward velocity of 20 m/s.V = 2 a tt = V/a = 20/g = 20/9.8 = 2.041 secondst = 2 x the gravity-drop = 4.082 seconds (rounded)
What is the boy throw a stone vertically upward with an initial velocity of 6.0 ms?
The boy throws a stone vertically upward with an initial velocity of 6.0 m/s, meaning the stone is moving against gravity. It will reach a maximum height and then fall back down due to gravity. The stone will eventually return to the ground after reaching its highest point.
If a projectile is fired straight up in the air at a speed of 30ms the total time to return to its starting point is about?
Assuming no air resistance, the time it takes for the projectile to return to its starting point is twice the time it takes to reach the highest point of its trajectory. The time to reach the highest point can be calculated using the equation: time = initial velocity / acceleration due to gravity. Therefore, the total time for the projectile to return would be around 6 seconds.
How do you return an object in java?
With the command return, followed by an object variable. In the method header, you have to declare the return type as the class of the object.
A ball is thrown upward with an initial velocity of 20 meter per second how long is the ball in the air?
Assuming no air resistance or external forces, the ball will be in the air for the same amount of time it takes for it to return to the same height it was thrown from. The time the ball will be in the air can be calculated using the kinematic equation: time = (2 * initial velocity) / gravitational acceleration. So, in this case, the ball will be in the air for approximately 4 seconds.
