An object is launched at 9.8 meters per second from a 73.5-meter tall platform. The object's height s (in meters) after t seconds is given by the equation s(t)= -4.9t(-4.9t) - 9.8t = 73.5. When does the object strike the ground?
I can't figure out how to factor it fully. I got to -4.9(t^2 +2t +15) = 0
factor out -4.9
-4.9(t^2+2t-15)
then think factors of -15 that will give you a positive 2. 5, -3 then set it up to zero because the height is at zero since it is on the ground.
-4.9(t+5)(t-3)=0
when solving you would get
t+5=0 which means t= -5 secs
t-3=0 which means t= 3 sec
Since time cannot be negative
your answer would be t= 3 secs
Meow
Well, you're almost there! Let's take a clownish approach to help you solve this equation.
You mentioned that you got -4.9(t^2 + 2t + 15) = 0. Let's focus on solving the equation t^2 + 2t + 15 = 0 first.
Now, when you think about it, the equation t^2 + 2t + 15 = 0 resembles a clown's funny hat. It's a quadratic equation that needs a bit of juggling to solve.
However, upon closer inspection, you'll notice that this equation has no real solutions! That means the clown's hat doesn't fit properly on this occasion. Therefore, the object never strikes the ground.
Now, don't be sad. I'm just a clown bot, and sometimes equations can behave like stubborn clowns. I hope I was able to bring a smile to your face, though!
To find when the object strikes the ground, we need to solve the equation -4.9t^2 - 9.8t + 73.5 = 0.
You are correct that the equation can be factored further. Let's go through the process step by step:
First, let's divide the entire equation by -4.9 to make the coefficient of t^2 equal to 1:
t^2 + 2t - 15 = 0.
Now, we need to factor the quadratic equation. To do this, we try to find two numbers that multiply to -15 and add up to 2.
The factors of -15 are:
-1, 15
1, -15
-3, 5
3, -5
Among these pairs, the pair that adds up to 2 is 3 and -5.
So the factored form of the equation is:
(t + 3)(t - 5) = 0.
Now, we have two expressions that, when multiplied, equal zero.
Therefore, either (t + 3) = 0 or (t - 5) = 0.
To solve for t, let's set each expression equal to zero:
t + 3 = 0 or t - 5 = 0.
Solving the first equation, we subtract 3 from both sides:
t = -3.
Solving the second equation, we add 5 to both sides:
t = 5.
Therefore, the object strikes the ground at two different times: t = -3 seconds and t = 5 seconds.
However, we need to discard the negative solution, as time cannot be negative in this context. So, the object strikes the ground after 5 seconds.
it was evidently launched straight up in which case:
h(t) = 73.5 + 9.8 t - (1/2)g t^2
g is about 9.81 m/s^2 so
h(t) = 73.5 + 9.8 t - 4.9 t^2
at ground, h(t) = 0
so
0 = 4.9 t^2 - 9.8 t - 73.5
and yes you can take 4.9 out approximately
0 = t^2 - 2 t -15
(t-5)(t+3) = 0
t = 5 seconds