The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21.5 m/s is
h = 3 + 21.5t − 4.9t2
after t seconds. (Round your answers to two decimal places.)
a. When does it hit the ground?
I set the equation equal to 0, yet I am not sure what to do since there is a number 3 in the equation, usually it's easy to simplify.
you are solving
4.9t^2 - 21.5t - 3 = 0 , after multiplying each term by -1
You must have learned the quadratic equation to be solving these type of questions, go ahead.
To determine when the projectile hits the ground, we need to find the value of t when the height (h) is equal to zero.
Given the equation: h = 3 + 21.5t - 4.9t^2
Setting h = 0, we get:
0 = 3 + 21.5t - 4.9t^2
Rearranging the equation, we have:
4.9t^2 - 21.5t - 3 = 0
This is a quadratic equation in the form of at^2 + bt + c = 0, where a = 4.9, b = -21.5, and c = -3.
To solve this equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values, we get:
t = (-(-21.5) ± √((-21.5)^2 - 4(4.9)(-3))) / (2(4.9))
Simplifying further:
t = (21.5 ± √(462.25 + 58.8)) / 9.8
t = (21.5 ± √521.05) / 9.8
Using a calculator to evaluate the square root, we have:
t ≈ (21.5 ± 22.83) / 9.8
t ≈ (21.5 + 22.83) / 9.8 or t ≈ (21.5 - 22.83) / 9.8
t ≈ 44.33 / 9.8 or t ≈ -1.33 / 9.8
t ≈ 4.53 or t ≈ -0.14
Since time cannot be negative in this context, we discard the negative solution.
Therefore, the projectile will hit the ground approximately at t ≈ 4.53 seconds.
To find the time when the projectile hits the ground, we need to set the height equation h = 3 + 21.5t - 4.9t^2 equal to 0.
Substituting h = 0, we get:
0 = 3 + 21.5t - 4.9t^2
Now, let's solve this quadratic equation for t using the quadratic formula. The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our case, we have:
a = -4.9, b = 21.5, and c = 3.
Plugging in these values, we get:
t = (-21.5 ± √(21.5^2 - 4(-4.9)(3))) / (2(-4.9))
Simplifying further:
t = (-21.5 ± √(462.25 + 59.28)) / (-9.8)
t = (-21.5 ± √521.53) / (-9.8)
Now, we have two possible solutions: t1 and t2.
t1 = (-21.5 + √521.53) / (-9.8)
t2 = (-21.5 - √521.53) / (-9.8)
Using a calculator, we find:
t1 ≈ 2.15 seconds
t2 ≈ -0.38 seconds
Since time cannot be negative, the projectile hits the ground approximately 2.15 seconds after it was shot upward.