A stone is thrown vertically upward with an initial speed u.if is the acceleration due to gravity,at what time will the stone return to the starting point
To find the time at which the stone returns to the starting point, we can use the equation of motion for vertical motion given by:
y = ut + (1/2)gt^2
Where:
y = displacement (change in height)
u = initial velocity
g = acceleration due to gravity
t = time
Since the stone is thrown vertically upward, the initial velocity (u) will be positive, and the acceleration due to gravity (g) will be negative.
At the starting point, the displacement (y) is zero, so we can set the equation equal to zero:
0 = ut - (1/2)gt^2
Rearranging the equation gives us:
(1/2)gt^2 = ut
Dividing both sides of the equation by t:
(1/2)gt = u
Now, we can solve for t:
t = (2u) / g
Therefore, the time at which the stone returns to the starting point is given by t = (2u) / g.
To find the time it takes for the stone to return to its starting point, we can use the kinematic equation for vertical motion:
s = ut + (1/2)gt²
Where:
- s is the displacement (change in position), which in this case is zero since the stone returns to its starting point.
- u is the initial velocity, given as u (upward).
- g is the acceleration due to gravity, assumed to be a positive value.
- t is the time.
Since the stone reaches back to its starting point, the displacement is zero. Therefore, we can write the equation as:
0 = ut + (1/2)gt²
Simplifying the equation, we get:
(1/2)gt² = -ut
Now, let's solve this equation for t:
(1/2)gt² + ut = 0
t²(g/2) + ut = 0
t²(g/2) + ut + (u²/g)(g/2) = (u²/g)(g/2)
Now, we have a quadratic equation in terms of t. Let's solve it using the quadratic formula:
t = (-u ± √(u² - 4(g/2)(u²/g)(g/2))) / (2(g/2))
Simplifying further:
t = (-u ± √(u² - 2ug(u²/g))) / g
t = (-u ± √(u² - 2u²)) / g
t = (-u ± √(u²(1 - 2))) / g
t = (-u ± √(-u²)) / g
Since time cannot be negative in this context, we discard the negative solution:
t = (-u + √(-u²)) / g
Now, it is important to note that the square root of a negative number does not have a real solution. Therefore, the stone will never return to its starting point if thrown vertically upward.