using the quadratic equation on a 2 dimensional projectile motion problem...
i am solving for tf and i have it worked all the way down to:
-100m(tf)-100m(tf) / -10m/s^2(tf)^2
or
-200m(tf) / -10m/s^2(tf)^2
is the answer 20 seconds? algebra is not my strong suit =(
What is the problem here. Is it something fired at some speed at some angle?
yes.. it's the "brothers of mercy" problem.
a plane is flying at 40 m/s, descending at an angle of 37 degrees to the horizontal. it is 100 m in the air and it is trying to drop a package to a man at R distance. what is R?
to find R, i needed to solve for time. that's where the quadratic came from
initial speed down = 40 sin 37
= 24.1 m/s
falls 100 meters, get speed down
v = 24.1 + 9.8 t
distance down
100 = 24.1 T + 4.9 T^2
4.9 T^2 + 24.1 T - 100 = 0
T = 2.68 s
now horizontal speed constant for 2.68 seconds
d = (40cos37)2.68
= 85.6 meters
To solve the quadratic equation and find the value of tf, let's simplify the expression you provided:
-200m(tf) / -10m/s^2(tf)^2
We can cancel out the common factor of "-10m" in the numerator and denominator, which gives us:
20 (tf) / (tf)^2
Now, we can rewrite the equation:
20 / tf
To solve for tf, we need to set the equation equal to zero. So, our equation is:
20 / tf = 0
To determine the value of tf, we need to find the value that makes the equation equal to zero. In this case, the denominator (tf) can't equal zero since division by zero is undefined. Therefore, we set the numerator equal to zero:
20 = 0
Since there is no value of tf that satisfies this equation, we conclude that there is no solution for tf in this quadratic equation. Therefore, we can't determine the value of tf based on the information provided.
It's worth noting that the quadratic equation is generally used to solve for the unknowns in equations that involve the square or square root of variables. In projectile motion, the quadratic equation is typically used to find the time of flight or the maximum height reached by the projectile. However, in this particular case, it seems like there may be a mistake in the derivation of the equation, as it does not lead to a meaningful solution.