A cannon ball is shot out of a cannon buried 2m in the ground. The height of the cannonball can be approximated by the equation h=-5t^2+35t-2 where is the height of the cannonball in meters and t is the time in seconds.
a)how long will it take the cannonball to reach ground level, to the nearest tenth of a second
b)how long is the cannonball in the air to the nearest tenth of a second
c)find the maximum height of the cannonball and the time it takes to reach this height.
what steps would i need to do this
To reach the ground, the height would be zero.
0 = -5t^2 + 35t - 2
You need to use the quadratic equation to solve for t, because this is not factorable.
b) Time up = time down
c) -b/2a will give you the max. height. -35/-10 = 3.5 meters
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To find the answers, you can follow these steps:
a) To determine when the cannonball reaches ground level, set the height equation h equal to 0 and solve for t:
-5t^2 + 35t - 2 = 0
You can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -5, b = 35, and c = -2. Solve for t, and round the answer to the nearest tenth of a second.
b) The time the cannonball is in the air is the time it takes to reach the ground from the initial launch. Therefore, it is the same as the answer to part a.
c) To find the maximum height, you can determine the vertex of the quadratic equation. The vertex form of a quadratic equation is given by h = a(t - t0)^2 + h0, where (t0, h0) is the vertex. In this case, a = -5.
The t-coordinate of the vertex can be found using the formula t0 = -b / (2a). Substitute the values of a and b, and calculate t0.
Finally, substitute the value of t0 into the original equation h = -5t^2 + 35t - 2 to find the maximum height.
To solve these problems related to the height of the cannonball, we will use the given equation h = -5t^2 + 35t - 2, where h is the height of the cannonball in meters and t is the time in seconds.
a) To find the time it takes for the cannonball to reach ground level (where the height is 0), we can set h equal to 0 and solve for t.
-5t^2 + 35t - 2 = 0
To solve this quadratic equation, you can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = -5, b = 35, and c = -2. Plugging these values into the quadratic formula will give you two possible solutions for t. Since we are looking for the time it takes for the cannonball to hit the ground, we only need the positive solution.
b) To find the total time the cannonball is in the air, we can find the time it takes to reach the ground and double it. This is because the time it takes for the cannonball to go up and come back down again is the same. So, multiply the positive solution obtained in part a by 2.
c) To find the maximum height of the cannonball and the time it takes to reach this height, we can find the vertex of the parabolic equation. The equation of the vertex is given by t = -b / (2a), which will give you the time at which the cannonball reaches its maximum height. Plug this value of t back into the equation to find the corresponding height.
Here are the detailed steps for solving each part:
a) Substitute h = 0 into our original equation:
-5t^2 + 35t - 2 = 0
Then, use the quadratic formula:
t = (-35 ± √(35^2 - 4(-5)(-2))) / (2(-5))
Simplify and calculate to find the positive solution in seconds.
b) Multiply the positive solution found in part a by 2 to find the total air time of the cannonball.
c) Use the equation for the vertex of a parabola:
t = -b / (2a)
Plug the values of a and b from our equation into this formula to find the time at which the cannonball reaches its maximum height.
Then, substitute this time value back into the original equation to find the corresponding maximum height.
Remember to round your answers to the nearest tenth of a second, as specified in the question.