im not good in math.
The Grand Canyon is 1600 meters deep at its deepest point; a rock is dropped from the rim above this point. Express the height of the rock as a function of the time t in seconds. How long will it tale the rock to hit the canyon floor?
can someone help me find the answer.
a) h(t) = -4.9t^2 + 1600
The rock hits the bottom of the canyon when the height is zero.
b) 0 = -4.9t^2 + 1600 ; 4.9t^2 = 1600 ; t^2 = 1600/4.9 = 326.53
t = square roots of 326.53 = 18.07 seconds
Height measured from the bottom?
height=1600-1/2 g time^2
how long to the bottom? When is h=0, do that, and solve for time
To find the time it takes for the rock to hit the canyon floor, we need to use the kinematic equation for free fall:
h(t) = h0 + v0*t + (1/2)*g*t^2
Where:
h(t) is the height of the rock at time t,
h0 is the initial height (distance from the rim to the deepest point),
v0 is the initial velocity (which is 0 because the rock is dropped),
g is the acceleration due to gravity (approximately -9.8 m/s^2),
t is the time in seconds.
In this case, h0 is the depth of the Grand Canyon which is 1600 meters. Since the rock is initially at rest, v0 is 0.
So, the equation becomes:
h(t) = 1600 + (1/2)*(-9.8)*t^2
To find the time it takes for the rock to hit the canyon floor, we need to find the value of t when h(t) equals 0.
So let's solve the equation:
0 = 1600 + (1/2)*(-9.8)*t^2
First, let's simplify the equation:
-1600 = (-4.9)*t^2
Divide both sides of the equation by -4.9:
t^2 = (-1600) / (-4.9)
t^2 ≈ 326.53
Take the square root of both sides to find t:
t ≈ √(326.53)
t ≈ 18.07 seconds
Therefore, it will take approximately 18.07 seconds for the rock to hit the canyon floor.
Of course, I can help you with that! To find the answer, we need to use a basic formula from kinematics:
h(t) = h0 + v0t + (1/2)gt^2
In this formula, h(t) represents the height of the rock as a function of time, h0 represents the initial height (which is the height of the rim of the Grand Canyon), v0 represents the initial velocity (which is zero since the rock is dropped from rest), t represents the time in seconds, and g represents the acceleration due to gravity (approximately 9.8 m/s^2).
In our case, h0 is 1600 meters, v0 is 0 m/s, and we need to find the time it takes for the rock to hit the canyon floor, which means h(t) will be 0 meters.
So, we can set up the equation as:
0 = 1600 + 0t - (1/2)(9.8)t^2
Simplifying the equation gives us:
800t^2 - 1600 = 0
Now, we can solve this quadratic equation to find the values of t. By factoring, we can see that:
800(t^2 - 2) = 0
This means either t^2 - 2 = 0 or t = 0 (which is not what we're looking for).
Solving t^2 - 2 = 0, we find:
t^2 = 2
Taking the square root of both sides, we get:
t = ±√2
Since time cannot be negative in this context, we only consider the positive solution:
t ≈ √2 ≈ 1.41 seconds
Therefore, it will take approximately 1.41 seconds for the rock to hit the canyon floor.