an object is launched from ground level with an initial velocity of 400 feet per second. To the nearest tenth off a second, how long does it take the object to reach a height off 650 feet?
1.3 seconds
its 1.7 had it in a quiz
To find the time it takes for the object to reach a certain height, we can use the kinematic equation for vertical displacement:
h = v0 * t + (1/2) * a * t^2
where:
h = height
v0 = initial velocity
t = time
a = acceleration (in this case, due to gravity, it is -32.2 ft/s^2)
In this case, we're given:
h = 650 ft
v0 = 400 ft/s
a = -32.2 ft/s^2
Substituting these values into the equation, we get:
650 = 400 * t + (1/2) * (-32.2) * t^2
Rearranging the equation and simplifying, we have:
0 = -16.1t^2 + 400t - 650
Now, to find the time it takes for the object to reach a height of 650 feet, we need to solve this quadratic equation. We can do this by using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16.1, b = 400, and c = -650. Substituting these values into the quadratic formula, we have:
t = (-400 ± √(400^2 - 4(-16.1)(-650))) / (2(-16.1))
Simplifying the equation further:
t = (-400 ± √(160000 - 41728)) / (-32.2)
t = (-400 ± √(118272)) / (-32.2)
Now, we have two possible values for t:
t1 = (-400 + √(118272)) / (-32.2)
t2 = (-400 - √(118272)) / (-32.2)
Calculating t1 and t2 using a calculator, we get:
t1 ≈ 12.8 seconds
t2 ≈ -0.3 seconds
We disregard the negative value since time cannot be negative in this context.
Therefore, the object takes approximately 12.8 seconds (to the nearest tenth of a second) to reach a height of 650 feet.
h = Vo*t + 0.5g*t^2 = 650 Ft,
400t + 16t^2 = 650,
16t^2 + 400t -650 = 0,
Use Quadratic Formula to find t:
t = 1.53s.