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How do you find the launch velocity of a projectile that was fired straight up and came back down to the level it was launched at in 1.37 seconds?
the velocity at the start is the same as when it comes back down to the level. At the top velocity is zero and at the bottom the velocity is acceleration x time where time is 1.37/2 = 0.685s. Acceleration is that of gravity or 9.8 m/s/s, so projectile velocity is 9.8 x .685 = 6.7 m/s; in US system that is 22 ft/sec or 15 mph
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If at the Earth's surface a projectile is launched straight up at a speed of 9 km per sec to what height will it rise Ignore air resistance and the rotation of the Earth?
9 km/s = 9000m/s Gravity decreases the velocity of the object by 9.8 m/s each second. The velocity at the top is 0 m/s Equation 1: Velocity final = velocity initial - (9.8 m/s × time) Final velocity =0 m/s Initial velocity = 9000m/s 0 = 9000 - 9.8 t 9.8 t = 9000 t = 9000÷ 9.8 t = 918 seconds Average velocity = (9000 + 0 ) ÷ 2 =4500 m/s Height = average velocity × time Height = 4500 m/s ×918 seconds=4,131,000 meters = 4,131 Km. If you do not want to round, this equation will find the answer more accurately. (velocity final) 2 - (velocity initial) 2 = 2 × acceleration × distance m/s2 0 - 90002 = 2 × 9.8 × d d = 4,132,653.061 meters = 4,132.653061Km I do not know of any measuring tool that measures that precisely!
What is the average velocity between 12.00 and 15.00 seconds?
To find average velocity, you need to know the displacement. If you knew displacement, average velocity would be found by: V = Displacement / time
What would be the velocity of a deepwater wave with a wavelength of 50 meters and a period of 6.5 second?
The velocity of a deepwater wave can be calculated using the formula v = L/T, where v is the velocity, L is the wavelength (50 meters), and T is the period (6.5 seconds). Substituting the values gives v = 50 meters / 6.5 seconds ≈ 7.69 m/s.
How the angular velocity of binary stars are calculated?
By observation. The angular velocity can be derived from the period. If, for example, it takes a day (86,400 seconds) for a full revolution, then the angular velocity will be (2 x pi / 86400) radians per second.
What graph of acceleration vs time constant velocity?
When the acceleration of a particle is constant, the velocity will be increasing at a constant rate. This means that the velocity versus time graph will appear with a straight line "slanting up to the right" in the first quadrant. With time on the x-axis and velocity of the y-axis, as time increases, velocity will increase. That means the line will have a positive slope. The higher the (constant) acceleration, the greater the slope of the line. If we take just one example and mark equal units off on our axes, and then assign seconds along the x-axis and meters per second along the y-axis, we can plot a graph for an acceleration of, say, one meter per second per second. Start at (0,0) and at the end of one second, the velocity will be one m/sec. That point will be (1,1). After another second, the velocity will be 2 m/sec owing to that 1m/sec2 rate of acceleration, and that point will be (2,2). The slope of the line is 1, which is the rate of acceleration.
You throw a ball straight up with a velocity of 40 meters per second. What is the ball's velocity after 3 seconds?
You throw a ball straight up with a velocity of 40 meters per second. What is the ball's velocity after 3 seconds?
In 0.5 seconds a projectile goes from 0 to 300 ms. what is the acceleration of the projectile?
The acceleration of the projectile can be calculated using the formula: acceleration = change in velocity / time taken. In this case, the change in velocity is 300 m/s (final velocity) - 0 m/s (initial velocity) = 300 m/s. The time taken is 0.5 seconds. Therefore, the acceleration would be 300 m/s / 0.5 s = 600 m/s².
Does motion of seconds hand of a watch represent uniform velocity?
No, the motion of the seconds hand of a watch is not an example of uniform velocity. The seconds hand moves in a circular motion at a constant speed, rather than moving in a straight line at a constant velocity. Uniform velocity refers to motion in a straight line with a constant speed.
You throw a ball straight up with a velocity of 40 meters per second. What is the ball's velocity after 5 seconds?
The velocity of the ball will be -30 m/s (downward) after 5 seconds due to gravity.
You throw a ball straight up with a velocity of 40 meters per second What is the ball's velocity after 5 seconds?
it is 10 meters per second straight down
A projectile is thrown with an initial velocity which has a horizontal component of 4 m s What will be its horizontal speed after 3s?
The horizontal speed of the projectile remains constant as there is no force acting in the horizontal direction to change it. Therefore, the horizontal speed of the projectile after 3 seconds will remain at 4 m/s.
If at the Earth's surface a projectile is launched straight up at a speed of 9 km per sec to what height will it rise Ignore air resistance and the rotation of the Earth?
9 km/s = 9000m/s Gravity decreases the velocity of the object by 9.8 m/s each second. The velocity at the top is 0 m/s Equation 1: Velocity final = velocity initial - (9.8 m/s × time) Final velocity =0 m/s Initial velocity = 9000m/s 0 = 9000 - 9.8 t 9.8 t = 9000 t = 9000÷ 9.8 t = 918 seconds Average velocity = (9000 + 0 ) ÷ 2 =4500 m/s Height = average velocity × time Height = 4500 m/s ×918 seconds=4,131,000 meters = 4,131 Km. If you do not want to round, this equation will find the answer more accurately. (velocity final) 2 - (velocity initial) 2 = 2 × acceleration × distance m/s2 0 - 90002 = 2 × 9.8 × d d = 4,132,653.061 meters = 4,132.653061Km I do not know of any measuring tool that measures that precisely!
In the oblique projection a projectile reaches the maximum height in 8 seconds it is time of flight is?
If a projectile takes 8 seconds to reach its maximum height, it will take another 8 seconds to return to its original elevation. Presuming it is lauched from flat ground and returns to the ground, its total time in flight would be 16 seconds. If it is launched from a hill, or at a hill, more information would be needed.
In 0.5 seconds a projectile goes from 0 to 300 meter per second What is the acceleration of the projectile?
The acceleration of the projectile can be calculated using the formula: acceleration = (final velocity - initial velocity) / time Plugging in the values: acceleration = (300 m/s - 0 m/s) / 0.5 s acceleration = 600 m/s^2 Therefore, the acceleration of the projectile is 600 meters per second squared.
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.
What is the velocity if you travel 882 ft in 5 seconds?
Assuming that you travel 882 feet in a straight line, the average velocity is 882.5 = 176.4 feet per second.
If a gun is fired straight upward and the bullet leaves the gun at 100ft per second after 2 seconds the velocity of the bullet is 36 feet per second what is the velocity after 3 seconds?
Since the bullet is fired straight upward, it will be decelerating due to gravity acting in the opposite direction of its initial velocity. Thus, at t = 3 seconds, the velocity will be less than 36ft per second but still positive (as it's moving upward). To calculate the exact velocity, you would need to use the kinematic equation for velocity in one dimension.
