A rock is thrown upward with a speed of 48 feet per second from the edge of a cliff 400 feet above the ground. What is the speed of the rock when it hits the ground? Use acceleration due to gravity as -32 feet per second squared and approximate your answer to 3 decimal places.
I don't understand exactly what this question is asking. I know velocity is the integral of acceleration, but how would I get the acceleration equation?
a(t) = -32 ft/sec^2 , given
v(t) = -32t + c
when t = 0, v(0) = 48 ft/sec, given
48 = 0 + c ---> c = 48
v(t) = -31t + 48
s(t) = -16t^2 + 48t + k
when t = 0 , s = 400 ft
s(t) = -16t^2 + 48t + 400
after doing a few of those you should be able to go directly to my last equation.
when it hits the ground, s(t) = 0
-16t^2 + 48t + 400 = 0
divide each term by -8
2t^2 - 6t - 50 = 0
use the quadratic equation formula to solve for t, one of the answers should be negative.
Use the positive root only
To solve this problem, you can use the kinematic equation for accelerated motion. The equation you mentioned, "velocity is the integral of acceleration," is correct, but in this case, we can simplify the problem by assuming constant acceleration due to gravity.
The kinematic equation for calculating the final velocity (v) of an object in free fall is:
v^2 = u^2 + 2as
where:
v = final velocity (what we want to find)
u = initial velocity (given as 48 feet per second, in the upward direction)
a = acceleration (given as -32 feet per second squared, in the downward direction)
s = displacement or distance covered (the distance from the cliff edge to the ground, given as 400 feet)
Now, substituting the given values into the equation, we have:
v^2 = (48)^2 + 2(-32)(400)
Simplifying,
v^2 = 2304 - 25600
v^2 = -23296
Since velocity cannot be negative, we only consider the positive value of v:
v = √(-23296)
Calculating this square root gives us an imaginary number because the value under the square root is negative. This indicates that the rock will not hit the ground.
Therefore, based on the given information, the rock will not hit the ground.
To understand the question, let's break it down step by step.
The initial velocity of the rock is 48 feet per second. It is thrown upward from the edge of a cliff that is 400 feet high from the ground. This means that initially, the rock is at a height of 400 feet above the ground.
The question asks for the speed of the rock when it hits the ground. To find this, we need to determine the time it takes for the rock to hit the ground.
To do this, we can use the equation of motion:
y = y0 + v0t + (1/2)at^2
Where:
y = final position (in this case, 0 feet as it hits the ground)
y0 = initial position (400 feet)
v0 = initial velocity (48 feet per second)
t = time
a = acceleration due to gravity (-32 feet per second squared)
Since we want to find the time it takes for the rock to hit the ground, we set y = 0 and solve for t:
0 = 400 + 48t - 16t^2
Simplifying the equation, we have:
16t^2 - 48t - 400 = 0
Now, we can solve this quadratic equation using any suitable method such as factoring, completing the square, or using the quadratic formula. In this case, using the quadratic formula would be the most straightforward approach.
The quadratic formula is:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where, in this case, a = 16, b = -48, and c = -400.
Now we can substitute the values into the formula:
t = (-(-48) ± √((-48)^2 - 4 * 16 * (-400))) / (2 * 16)
t = (48 ± √(2304 + 25600)) / 32
t = (48 ± √(27904)) / 32
t = (48 ± 167.18) / 32
We have two possible solutions, one with the plus sign and one with the minus sign:
t1 = (48 + 167.18) / 32 ≈ 7.716 seconds
t2 = (48 - 167.18) / 32 ≈ -3.809 seconds
Since time cannot be negative in this context, we discard the negative solution.
So, the time it takes for the rock to hit the ground is approximately 7.716 seconds.
Now, to find the final speed of the rock when it hits the ground, we can use the equation:
v = v0 + at
Substituting the known values:
v = 48 + (-32) * 7.716
v ≈ 48 - 246.912
v ≈ -198.912
Since velocity cannot be negative in this context, we take the magnitude of the velocity:
v ≈ 198.912
Therefore, the speed of the rock when it hits the ground is approximately 198.912 feet per second.