A stone is drop from a height 19.6m above the ground while a second stone is simultaneously projected from the ground with sufficient velocity to enable it to accept 19.6m. When and where the stone will meet
dropped stone:
hf=hi+vi*t-4.9t^2
hf=19.6-4.9t^2
thrown stone:
Initial KE =final PE of max height
1/2 mvi^2=mg*19.6
vi=sqrt(2*9.8*19.6)
vi= 19.6m/s
hf=hi+vi*t-4.9t^2
hf=19.6*t-4.9t^2
but the hf is the same for each stone.
19.6*t-4.9t^2 =19.6-4.9t^2
they meet at time t=1second
hf=19.6*t-4.9t^2....
To find when and where the two stones will meet, we can use the equations of motion.
Let's start by finding the time it takes for the first stone to fall from a height of 19.6m.
Using the equation of motion:
h = ut + (1/2)gt^2
Where:
h = height (19.6m in this case)
u = initial velocity (0 m/s since it is dropped)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the equation:
19.6 = 0*t + (1/2)*9.8*t^2
Simplifying:
19.6 = 4.9t^2
Dividing by 4.9:
t^2 = 4
Taking the square root of both sides:
t = 2 seconds
So it takes 2 seconds for the first stone to reach the ground.
Now, let's find the time it takes for the second stone to reach a height of 19.6m.
Using the equation of motion:
h = ut + (1/2)gt^2
Where:
h = height (19.6m in this case)
u = initial velocity (unknown)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Since the stone is projected with sufficient velocity to reach 19.6m, we can assume that the final velocity is 0 when it reaches that height.
Applying the equation v^2 = u^2 + 2gh:
0 = u^2 + 2*9.8*19.6
Simplifying:
u^2 = -2*9.8*19.6
u^2 = -384.16
u = √(-384.16) (ignoring the negative value as it represents an imaginary number)
Calculating the square root:
u ≈ 19.60 m/s
Now, we can find the time using the equation:
t = (v - u)/g
Where:
v = final velocity (0 m/s)
u = initial velocity (19.60 m/s)
g = acceleration due to gravity (9.8 m/s^2)
t = (0 - 19.60)/9.8
t = -19.60/9.8
t ≈ -2 seconds (Ignoring the negative value as it represents a time before it was projected)
Based on the calculations above, the second stone reaches a height of 19.6m in approximately 2 seconds after it was projected.
Since both stones take the same amount of time to reach the ground and the second stone is projected when the first one is dropped, they meet at the ground level.
Therefore, the stones will meet at the ground level after approximately 2 seconds.
To find out when and where the two stones will meet, we can use kinematic equations of motion. Let's consider the motion of each stone separately.
For the stone dropped from a height of 19.6m above the ground, we can use the equation for freefall motion:
h = ut + (1/2)gt^2
where:
h = height (19.6m in this case)
u = initial velocity (0 m/s as the stone is dropped)
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Plugging in the values, the equation becomes:
19.6 = 0t + (1/2)(-9.8)t^2
39.2 = -4.9t^2
t^2 = -39.2/-4.9
t^2 = 8
t = √8
We discard the negative value for time, so t = √8 = 2.83 seconds.
Now, let's determine the position of the stone projected from the ground with sufficient velocity to reach a height of 19.6m. We can use the equation for projectile motion:
h = ut + (1/2)gt^2
Since the stone is projected, we need to find the initial velocity (u). We can use the formula for distance traveled in projectile motion:
h = (u^2 sin^2θ)/(2g)
where:
θ = angle of projection
u = initial velocity
In this case, the stone reaches a height of 19.6m when it comes to rest, which means its final velocity (v) is zero. We can use v = u + gt, where v = 0 and g = 9.8 m/s^2.
0 = u + (9.8)(2.83)
u = -9.8(2.83)
u = -27.67 m/s
Now that we have the initial velocity, we can use the equation for projectile motion:
19.6 = (-27.67)t + (1/2)(-9.8)t^2
19.6 = -27.67t - 4.9t^2
Rearranging the equation and setting it equal to zero:
4.9t^2 + 27.67t - 19.6 = 0
We can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
where a = 4.9, b = 27.67, and c = -19.6.
Plugging in the values:
t = (-27.67 ± √(27.67^2 - 4(4.9)(-19.6)))/(2(4.9))
Calculating the values inside the square root:
t = (-27.67 ± √(765.0489 + 385.28))/9.8
t = (-27.67 ± √(1150.3289))/9.8
t = (-27.67 ± 33.93)/9.8
Simplifying:
t = (6.26)/9.8 or t = (-61.6)/9.8
We discard the negative value for time, so t = 6.26/9.8 = 0.638 seconds.
So, the stones will meet after approximately 0.638 seconds. To find the height at which they meet, we can use the equation for the stone dropped from the height of 19.6m:
h = 0t + (1/2)(-9.8)t^2
h = (1/2)(-9.8)(0.638)^2
h = -4.9(0.406)
h = -1.988
The negative sign indicates that the height is below the ground level. So, the stones will meet at a height of approximately 1.988m below the ground level.