A rock is thrown vertically up from the ground with a velocity of 24 meters per second, and it reaches a height of 2+24t-4.9tsquare after t secounds.
A. How many seconds after the rock is thrown will it reach its maximum height?
B.What is the maximum height the rock will reach, in meters?
C. How many seconds after the rock is thrown will it hit the ground?
It will reach the max height after 2.45 secs.
You can find the max height by differentiation and setting it to zero. ie. 24 -9.8 t = 0 and solve for t to be 2.45 secs.
Then use 2.45 in the equation to find the max height to be 2 + 24 *2.45 + 4.9 (2.45)(2.45) = 90.21 meters.
It will hit the ground in 2.45 * 2 secs = 4.90 secs
To find the answer to part A, we need to determine when the rock reaches its maximum height. This occurs when the velocity of the rock becomes zero, as it changes direction from going up to coming back down.
In the given problem, the velocity is given as 24 meters per second. We can set up the equation 24 - 9.8t = 0, where t represents time in seconds. This equation comes from the fact that the acceleration due to gravity is -9.8 m/s^2 (negative because it is acting in the opposite direction to the motion). Solving for t gives us t = 24/9.8, which is approximately 2.45 seconds.
Therefore, the rock will reach its maximum height after 2.45 seconds.
To find the answer to part B, we substitute the value of t = 2.45 into the given equation: 2 + 24(2.45) - 4.9(2.45)^2. Evaluating this expression gives us a maximum height of approximately 90.21 meters.
Lastly, to find the answer to part C, we can consider when the rock hits the ground. In this case, the height is equal to zero, so we can set the given equation 2 + 24t - 4.9t^2 = 0. We can solve this quadratic equation to find the value of t when the height is zero. Using the quadratic formula or factoring, we find two values of t: t = 0 and t = 4.9 seconds.
Since t = 0 represents the initial time when the rock is thrown, we are only interested in t = 4.9 seconds as the time it takes for the rock to hit the ground.
Therefore, the rock will hit the ground in approximately 4.9 seconds.
To find the maximum height the rock will reach, we need to differentiate the equation with respect to time and set it equal to zero.
The equation is given as:
h(t) = 2 + 24t - 4.9t^2
Differentiating the equation with respect to time (t), we get:
dh(t)/dt = 24 - 9.8t
Setting this derivative equal to zero, we have:
24 - 9.8t = 0
Solving for t:
9.8t = 24
t = 24 / 9.8
t ≈ 2.45 seconds
Therefore, the rock will reach its maximum height approximately 2.45 seconds after it is thrown.
To find the maximum height, substitute this value of t back into the equation:
h_max = 2 + 24(2.45) - 4.9(2.45)^2
h_max = 2 + 58.8 - 4.9(6.0025)
h_max = 2 + 58.8 - 29.41325
h_max ≈ 90.21 meters
So, the maximum height the rock will reach is approximately 90.21 meters.
To find how many seconds after the rock is thrown it will hit the ground, we need to set the height equation equal to zero and solve for t:
2 + 24t - 4.9t^2 = 0
This is a quadratic equation, and we can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -4.9, b = 24, and c = 2.
Plugging in these values, we get:
t = (-24 ± √(24^2 - 4(-4.9)(2))) / (2(-4.9))
Simplifying the equation further, we have:
t = (-24 ± √(576 + 39.2)) / (-9.8)
t = (-24 ± √615.2) / (-9.8)
Taking the positive root, we have:
t ≈ (-24 + √615.2) / (-9.8)
t ≈ -4.9 seconds (which is not meaningful in this context) or t ≈ 4.90 seconds
Therefore, the rock will hit the ground approximately 4.90 seconds after it is thrown.