If an object is thrown downward with an initial velocity of
v0, then the distance it travels is given by
s=4.9t sqrt 2 +v v0t. An object is thrown downward from an airplane 100 m from the ground, with an initial velocity of 6
m/sec. How long does it take for the object to reach the ground?
your equation should be
s = -4t^2 - 6t + 100
you want to know when s = 0
4t^2 + 6t - 100 = 0
2t^2 + 3t - 50 = 0
t = (-3 ±√409)/4
= appr 4.3 seconds or a negative, which we will reject
To find out how long it takes for the object to reach the ground, we need to find the value of 't' for which the distance traveled 's' is equal to the height from the airplane to the ground, which is 100 m.
The given equation is:
s = 4.9t√2 + v₀t
We know that s is equal to 100 m and v₀ is equal to -6 m/s (since the object is thrown downward). Substituting these values into the equation, we have:
100 = 4.9t√2 - 6t
To solve for 't', we need to rearrange the equation. Adding 6t to both sides, we get:
4.9t√2 + 6t = 100
Now, we'll simplify the equation by multiplying both sides by √2 to get rid of the radical:
[4.9√2]t + [6√2]t = 100√2
Next, we'll express the coefficients as single terms:
(4.9√2 + 6√2)t = 100√2
Combine like terms:
(4.9 + 6)√2t = 100√2
Simplify:
10.9√2t = 100√2
Divide both sides by 10.9√2:
t = 100√2 / 10.9√2
The square root of 2 on numerator and denominator cancels out:
t = 100 / 10.9
Now, divide 100 by 10.9:
t ≈ 9.174
Therefore, it takes approximately 9.174 seconds for the object to reach the ground.