a stone is thrown vertically upwards with an initial velocity of 14m/s .neglecting air resistence , find time taken before it reaches the ground .
it hits the ground when the height is zero, right? So just solve
14t - 4.9t^2 = 0
assuming it was thrown from the ground ...
I don't under stand what you wrote it is confusing
To find the time taken before the stone reaches the ground, we can use the equation of motion for vertical motion with constant acceleration. In this case, the acceleration is due to gravity and is equal to -9.8 m/s². We can use the following equation:
h = ut + (1/2)gt²
Where:
h = height or displacement
u = initial velocity
t = time
g = acceleration due to gravity
Since the stone is thrown vertically upwards, the height h is equal to zero when it reaches the ground. Also, the initial velocity u is positive (+14 m/s) as it is thrown upwards. Therefore, the equation becomes:
0 = 14t - (1/2)(9.8)t²
Simplifying the equation:
0 = 14t - 4.9t²
Rearranging the equation:
4.9t² - 14t = 0
Now, we can solve this quadratic equation for time (t). We can factor out t:
t(4.9t - 14) = 0
Set each factor equal to zero:
t = 0 or 4.9t - 14 = 0
For the stone to reach the ground, t cannot be zero. Therefore, we solve the second equation:
4.9t - 14 = 0
Adding 14 to both sides:
4.9t = 14
Dividing both sides by 4.9:
t = 14 / 4.9 ≈ 2.86 seconds
So, the time taken before the stone reaches the ground is approximately 2.86 seconds.