An object is projected vertically upwards and returns to the point of projection 12s later. Ignore air resistance and determine the velocity with which it was projected?
Vf=Vi +at
Vf=0, Vi=?, a=-9.8 remeber time taken to go up is equal to the time taken to go back to the point of projectile.. THEREFORE t=6
Vf=Vi +at
(0)=Vi +(-9.8)(6)
Vi=58.8m/s
***[OR]***
Vf=Vi +at
Vf=-Vi, Vi=?, a=-9.8, t=12
(-Vi)=Vi +(-9.8)(12)
-2Vi= -117.6
Vi=58.8m/s
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It corect
To determine the velocity with which the object was projected, we can use the equations of motion.
The key information provided in the question is that the total time taken for the object to go up and come back down is 12 seconds. This means that the time taken for the object to reach its highest point (the peak of its trajectory) is half of 12 seconds, which is 6 seconds.
We can use the equation for displacement to find the time it takes for the object to reach its highest point:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity
t = time taken
a = acceleration (which is equal to the acceleration due to gravity, g = -9.8 m/s^2)
At the highest point, the displacement is zero because the object has reached the peak of its trajectory. Substituting the provided values into the equation:
0 = u(6) + (1/2)(-9.8)(6)^2
Simplifying this equation, we get:
0 = 6u - 176.4
Rearranging, we find:
6u = 176.4
Dividing both sides by 6:
u = 29.4 m/s
Therefore, the velocity with which the object was projected vertically upwards is 29.4 m/s.