A stone is dropped from a high altitude and 3 s later another stone is projected downward with a speed of 150 ft/s. When and where will the second stone overtake the first? Ans. (a) 5.70 s (b) 520 ft
How did u get the 5.7
h(1st stone)= h(2nd stone)
1/2 (g)(t^2) = Vot - 1/2 (g)(t^2)
let x= time of 2nd stone to reach 1st stone
1/2 (32.22)(x + 3) = 150(x) + 1/2 (32.22)(x^2)
x=2.7
total time=3+2.7=5.7s
height =1/2 (32.22)(5.7)^2= 523.4ft
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To find when and where the second stone overtakes the first, we need to calculate the time and the position of each stone at that moment.
Let's start by finding the time it takes for the first stone to fall after the second stone is projected downward. We know that the first stone has a headstart of 3 seconds. So, we can ignore the first 3 seconds and focus on finding when the second stone catches up.
First, let's find the time it takes for the second stone to fall:
Using the kinematic equation:
d = v0 * t + (1/2) * a * t^2
Where:
d = displacement
v0 = initial velocity
t = time
a = acceleration due to gravity (approximated as -32 ft/s^2)
Since the second stone is projected downward, its initial velocity is 150 ft/s and the acceleration due to gravity is still -32 ft/s^2.
Let t2 be the time for the second stone to fall:
0 = (150 ft/s) * t2 + (1/2) * (-32 ft/s^2) * (t2)^2
Simplifying the equation:
16t2^2 - 150t2 = 0
t2(16t2 - 150) = 0
Using the zero product property, we have two options:
t2 = 0 (discarding it as we're looking for a positive time)
16t2 - 150 = 0
Solving the second equation:
16t2 = 150
t2 = 150/16
t2 ≈ 9.375 s
So, it takes approximately 9.375 seconds for the second stone to fall.
Now, let's find the position at that time for both stones. We first need to find the position of the first stone after 9.375 seconds. Since it was dropped from a high altitude, its initial velocity is 0 ft/s, and the acceleration due to gravity is -32 ft/s^2.
Let h1 be the position of the first stone after 9.375 seconds:
h1 = (1/2) * (-32 ft/s^2) * (9.375 s)^2
h1 = -120.94 ft
Therefore, after 9.375 seconds, the first stone is approximately 120.94 ft below its starting point.
Next, we need to find the position of the second stone after 9.375 seconds. Since its initial velocity is 150 ft/s and it's projected downward, the acceleration due to gravity is still -32 ft/s^2. Let h2 be the position of the second stone after 9.375 seconds:
h2 = (150 ft/s) * (9.375 s) + (1/2) * (-32 ft/s^2) * (9.375 s)^2
h2 = 1406.25 ft + (-480.47 ft)
h2 = 925.78 ft
Therefore, after 9.375 seconds, the second stone is approximately 925.78 ft below its starting point.
To find when the second stone overtakes the first, we need to find the time it takes for the second stone to catch up. Let t be that time:
t = 9.375 s + 3 s
t ≈ 12.375 s
Therefore, the second stone will overtake the first after approximately 12.375 seconds.
To find where the second stone overtakes the first, we need to calculate the position at that moment. Again, let h be the position at that moment:
h = (150 ft/s) * (12.375 s) + (1/2) * (-32 ft/s^2) * (12.375 s)^2
h = 1856.25 ft + (-752.41 ft)
h ≈ 1103.84 ft
Therefore, the second stone will overtake the first after approximately 12.375 seconds at a position approximately 1103.84 ft below the starting point.
To find out when and where the second stone overtakes the first stone, we need to calculate the time and position at which they meet. We can break down the problem into two parts - the motion of the first stone and the motion of the second stone.
Let's start by finding the time it takes for the first stone to fall. We can use the formula for the distance traveled by an object in free fall:
d1 = 1/2 * g * t^2
Where:
d1 = distance traveled by the first stone
g = acceleration due to gravity (32 ft/s^2)
t = time elapsed (in seconds)
Since the first stone is dropped, it starts at rest, so its initial velocity (u) is 0. Therefore, the equation becomes:
d1 = 1/2 * g * t^2
We don't know the distance traveled by the first stone, but we know it takes 3 seconds to fall. So, we have:
3 = 1/2 * 32 * (3^2)
3 = 1/2 * 32 * 9
3 = 144/2
3 = 72
So, the distance traveled by the first stone is 72 ft.
Now, let's consider the motion of the second stone. It is projected downward with an initial velocity (u2) of 150 ft/s. We can calculate the distance traveled by the second stone using the kinematic equation:
d2 = u2 * t + 1/2 * a * t^2
Where:
d2 = distance traveled by the second stone
u2 = initial velocity of the second stone
t = time elapsed (in seconds)
a = acceleration (which is equal to -g in this case, since the second stone is moving downward)
We want to find the time (t) and distance (d2) at which the second stone overtakes the first. At that point, the distance traveled by both stones will be equal. So, we have:
d1 = d2
72 = u2 * t + 1/2 * (-g) * t^2
Substituting the values we have:
72 = 150 * t + 1/2 * (-32) * t^2
Rearranging the equation:
0 = 16t^2 - 150t + 72
Solving this quadratic equation will give us the value of t when the second stone overtakes the first:
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 16, b = -150, and c = 72.
t = (-(-150) ± √((-150)^2 - 4 * 16 * 72)) / (2 * 16)
t = (150 ± √(22500 - 4608)) / 32
t = (150 ± √(17892)) / 32
Using a calculator, we find:
t ≈ 5.70 seconds
So, the second stone overtakes the first stone approximately 5.70 seconds after it was projected.
To find the distance covered by the second stone at that time, substitute the value of t back into the equation for d2:
d2 = 150 * t + 1/2 * (-g) * t^2
d2 = 150 * 5.7 + 1/2 * (-32) * (5.7^2)
Using a calculator, we find:
d2 ≈ 520 ft
Therefore, the second stone overtakes the first stone after approximately 5.70 seconds and at a distance of 520 ft.