An object is launched straight up into the air with an initial velocity of 40 m/s, from a height 30 m above the ground. Assuming that gravity pulls it down, changing its position by about 4.9 m/s2, after how many seconds will the object hit the ground?
To find out how many seconds it will take for the object to hit the ground, we can use the kinematic equation:
h = h0 + v0t - (1/2)gt^2
where:
h = final height (0 m, since it hits the ground)
h0 = initial height (30 m)
v0 = initial velocity (40 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
Plugging in the values, we have:
0 = 30 + 40t - (1/2)(-9.8)t^2
Simplifying the equation, we get:
4.9t^2 - 40t - 30 = 0
We can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where:
a = 4.9
b = -40
c = -30
Plugging in these values, we get:
t = (-(-40) ± √((-40)^2 - 4(4.9)(-30))) / (2(4.9))
Simplifying further:
t = (40 ± √(1600 + 588)) / 9.8
t = (40 ± √2188) / 9.8
Using a calculator, we can determine the values of t:
t1 ≈ 5.44 seconds
t2 ≈ -0.71 seconds
Since time cannot be negative, the object will hit the ground approximately 5.44 seconds after it was launched.
To find out how many seconds it will take for the object to hit the ground, we can use the kinematic equation for vertical motion:
h = h₀ + v₀t - (1/2)gt²
Where:
h is the height of the object at time t
h₀ is the initial height of the object (30 m in this case)
v₀ is the initial velocity of the object (40 m/s in this case)
g is the acceleration due to gravity (-9.8 m/s², assuming downward is positive)
t is the time in seconds
Since the object hits the ground when its height is zero, we can set h equal to zero and solve for t:
0 = 30 + 40t - (1/2)(-9.8)t²
Simplifying the equation:
0 = 30 + 40t + 4.9t²
Now, we have a quadratic equation. To solve it, we can either factor it, complete the square, or use the quadratic formula. Let's use the quadratic formula:
t = [-b ± √(b² - 4ac)] / (2a)
where a = 4.9, b = 40, and c = 30
Substituting the values:
t = [-40 ± √(40² - 4(4.9)(30))] / (2(4.9))
Simplifying further:
t = [-40 ± √(1600 - 588)] / 9.8
t = [-40 ± √(1012)] / 9.8
The expression inside the square root can be simplified as the square of (√4)(√253), which gives 2√253. Therefore:
t = [-40 ± 2√253] / 9.8
Now, we have two possible solutions for t:
t₁ = [-40 + 2√253] / 9.8
t₂ = [-40 - 2√253] / 9.8
These solutions represent the times at which the object reaches the ground. Since time cannot be negative, we can discard t₂. Therefore, the object hits the ground after approximately:
t ≈ [-40 + 2√253] / 9.8 seconds