on earth, in the absence of air, the rock in excersise 13 would reach a height of s=24t- 4.9t^2 meters in t seconds. How high would the rock go?

(exercise 13:a rock thrown vertically upward from the surface of the moon at a velocity of 24m/sec reaches aheight of s=24t-0.t^2.

Hey there! It seems like you're doing some physics exercises on the moon, how cool! So, let's see how high that rock on the moon can go.

According to the equation s = 24t - 4.9t^2, where s represents the height in meters and t represents time in seconds, we need to find the maximum height.

Now, since we're looking for the maximum height, we know that at the peak of the motion, the velocity becomes zero. So, let's find that moment by setting the equation equal to zero:

0 = 24t - 4.9t^2

By rearranging the equation, we get:

4.9t^2 - 24t = 0

Now we can solve for t by factoring out common terms:

t(4.9t - 24) = 0

This means that t = 0 or t = 24/4.9. Since t cannot be zero (otherwise the rock wouldn't have been thrown), we'll go with t = 24/4.9.

Plugging this value back into the original equation:

s = 24(24/4.9) - 4.9(24/4.9)^2

Doing the math, we find that the maximum height the rock would reach on the surface of the moon is approximately 116.49 meters.

So, there you have it! The rock would go up to about 116.49 meters high on the moon. Hope this helps, and keep rocking those exercises!

To find the maximum height that the rock would reach, we need to determine the value of t when the rock reaches its highest point. This occurs when the velocity of the rock becomes zero.

Given the equation for the height of the rock: s = 24t - 4.9t^2

To find the time at which the velocity becomes zero, we differentiate the equation with respect to t and set the derivative equal to zero:

s' = 24 - 9.8t

Setting s' equal to zero and solving for t:

24 - 9.8t = 0
9.8t = 24
t = 24 / 9.8
t ≈ 2.45 seconds

Substituting this value of t into the original equation, we can determine the maximum height:

s = 24t - 4.9t^2
s = 24(2.45) - 4.9(2.45)^2
s ≈ 58.8 - 29.26
s ≈ 29.54 meters

Therefore, the rock would reach a maximum height of approximately 29.54 meters.

To find the maximum height the rock would reach on Earth in the absence of air, we can use the given equation s = 24t - 4.9t^2, where s represents the height of the rock in meters and t represents the time in seconds.

To find the maximum height, we need to find the vertex of the parabolic equation. The vertex represents the maximum or minimum point of the parabola. In this case, since the rock is thrown vertically upward, the vertex represents the maximum height.

The formula to find the x-coordinate of the vertex is given by x = -b / (2a), where a and b are coefficients of the quadratic equation ax^2 + bx + c = 0. In our case, the equation is -4.9t^2 + 24t, so a = -4.9 and b = 24.

Using the formula, we can find the time when the rock reaches its maximum height:

t = -24 / (2 * -4.9)
t = -24 / -9.8
t = 2.45 seconds

Now that we have the time (t), we can substitute it back into the equation to find the maximum height (s):

s = 24t - 4.9t^2
s = 24 * 2.45 - 4.9 * (2.45)^2
s = 58.8 - 4.9 * 6.0025
s = 58.8 - 29.41375
s ≈ 29.39 meters

Therefore, the rock would reach a height of approximately 29.39 meters on Earth in the absence of air.

29.388meters

52 seconds