The flight of a ball hit from a tree that is 0.6m tall can be modelled by the function h(t)=-4.9t+6t+0.6, where h(t) is the height in the metres at time t seconds. How long will it take for the ball to hit the ground?
it hits the ground when the height is zero, right?
So just solve
-4.9t^2 + 6t + 0.6 = 0
using the quadratic formula is probably easiest.
Quadratic Formula -b±√(b²-4ac) /(2a)
a = -4.9
b= 6
c=0.6
-6±√6²-4(-4.9)(0.6) /(2)(-4.9)
= -6±√36+11.76 /-9.8
= -6±√47.76 /-9.8
= -6+ 6.9/ -9.8, -6- 6.9/ -9.8
x = -0.09, 1.31s
Therefore, it will take the ball 1.31 seconds to hit the ground.
*If the question specifies rounding to two decimal places,
x = 1.32s
Well, let's solve this problem...but first, let me grab my umbrella in case the ball hits me on its way down!
To find out when the ball hits the ground, we need to set h(t) equal to zero and solve for t. So, our equation is:
0 = -4.9t^2 + 6t + 0.6
Hmmm, seems like a job for quadratic equations! But I'm more of a comedian than a mathematician, so brace yourself for some wordplay!
Let's look at the equation. It's like a ballerina trying to find balance – she needs to find that point where she neither goes up nor comes down. In other words, we're searching for the value of t when the ball's height is absolutely zero.
To find that special t-value, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a. But fear not, I'll do the math for you!
Plugging in the values from our equation, we get: t = [ -6 ± √(6^2 - 4(-4.9)(0.6))] / (2 * -4.9)
Now, let me entertain you with some number crunching...
After doing the math, I have not one but TWO possible answers. Drumroll, please...
t ≈ 0.133 seconds or t ≈ 1.14 seconds
That means the ball will hit the ground either in a blink of an eye or just enough time for you to Google "funny clown jokes" and have a laugh before the ball lands.
Just remember, laughter is the best medicine, especially when dodging falling balls! Enjoy your day and stay safe!
To find the time it takes for the ball to hit the ground, we need to find when the height, h(t), is equal to zero.
The given function is h(t) = -4.9t^2 + 6t + 0.6.
Setting h(t) equal to zero, we have:
0 = -4.9t^2 + 6t + 0.6
Now, we can solve this equation to find the value of t.
We can first divide the entire equation by -0.1 to simplify it:
0 = 49t^2 - 60t - 6
Next, we can rearrange the equation to match the standard quadratic form:
49t^2 - 60t - 6 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 49, b = -60, and c = -6.
Now, substituting these values into the quadratic formula, we get:
t = (-(-60) ± √((-60)^2 - 4 * 49 * (-6))) / (2 * 49)
Simplifying further:
t = (60 ± √(3600 + 1176)) / 98
t = (60 ± √(4776)) / 98
Now, calculate the square root of 4776:
√(4776) ≈ 69.09
Plugging this value back into the equation, we have:
t = (60 ± 69.09) / 98
Now, the two possible solutions are:
t1 = (60 + 69.09) / 98 ≈ 1.3
t2 = (60 - 69.09) / 98 ≈ -0.9
Since time cannot be negative in this context, the only valid solution is t = 1.3 seconds.
Therefore, it will take approximately 1.3 seconds for the ball to hit the ground.
To find the time it takes for the ball to hit the ground, we need to determine when the height of the ball, represented by the function h(t), reaches 0.
The equation for the height of the ball is h(t) = -4.9t^2 + 6t + 0.6. We set h(t) = 0 and solve for t.
0 = -4.9t^2 + 6t + 0.6
To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -4.9, b = 6, and c = 0.6:
t = (-6 ± √(6^2 - 4(-4.9)(0.6))) / (2(-4.9))
t = (-6 ± √(36 + 11.76)) / (-9.8)
t = (-6 ± √47.76) / (-9.8)
Now, we have two solutions for t, but we are only interested in the positive one since time cannot be negative in this context.
t = (-6 + √47.76) / (-9.8) ≈ 1.18 seconds
Therefore, it will take approximately 1.18 seconds for the ball to hit the ground.