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)
After just over three and a quarter seconds.
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
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.
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.
it would be slower
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.
Depending on the value used for the acceleration due to gravity (9.8 ms-2 or 9.81 ms-2), the ball will go up to 11.48 m or 11.47 m (approx). It will be in the air for a total of 3.06 seconds.
Assumptions: 1) You throw it straight up. No angle, nothing. 2) Answers.com removed your punctuation, thus you meant initial speed of 39 m/s 3) No air resistance 4) Your "throw" begins at the ground (meaning you're not counting the height of the person) Answer = 39 m/s Reasoning: With nothing but the acceleration of gravity to slow it, it will slow to a stop, then accelerate and return along the same path. With no resistance, it will accelerate to the same velocity as its initial velocity by the time it reaches the point of origin. Fun Fact: Feel free to try this. I recommend using a hackey-sack and laying on your back while you throw upward.
Pulse rate is either set up by the operator or the equipment designer. Range of some detected object is determined by the speed of light and the distance of the object causing the return signal pulse, using the formula: distance= velocity of light multiplied by half the time delay for the return echo.
When you pick an object and do not return it, in probability it is termed "without replacement".
The duration of Return of the Ewok is 1440.0 seconds.
The duration of Return to Glennascaul is 1380.0 seconds.