An object with mass 8.5kg is moving vertically upward with a velocity if 35m/s at a height of 100m above the ground.

Find the time the object takes to reach the ground

Thanks!!

Kinematics equation:

Δx = vi*Δt + (1/2)aΔt²

Here
Δt = elapsed time, in seconds
vi=initial velocity = +35 m/s
a=acceleration due to gravity = -9.8 m/s²
Δx=displacement = -100 m

-100=35Δt + (-9.8/2)Δt²
Solve by quadratic formula for Δt
and reject negative root:
Δt
= (5√65+25)/7
= 9.88 s (approx.)

To find the time it takes for the object to reach the ground, we can use the equation of motion in the vertical direction:

h = vi * t + (1/2) * g * t^2

Where:
h = height above the ground (100m)
vi = initial velocity (35m/s)
g = acceleration due to gravity (-9.8m/s^2)
t = time taken to reach the ground (unknown)

Now, let's solve for t:

Rearrange the equation to isolate t:

0 = (1/2) * g * t^2 + vi * t - h

Now, plug in the values:

0 = (1/2) * (-9.8) * t^2 + 35 * t - 100

This equation is quadratic, so we can solve for t using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case:
a = (1/2) * (-9.8) = -4.9
b = 35
c = -100

After substituting the values into the quadratic formula, we get:

t = (-35 ± √(35^2 - 4 * (-4.9) * (-100))) / (2 * (-4.9))

Simplifying further gives us:

t = (-35 ± √(1225 + 1960))/(-9.8)

Now, calculate the discriminant inside the square root:

t = (-35 ± √(3185))/(-9.8)

Since we're looking for the time it takes for the object to reach the ground, we only consider the positive root:

t = (-35 + √(3185))/(-9.8)

So, the time it takes for the object to reach the ground is approximately:

t = 5.81 seconds (rounded to two decimal places)