If a projectile is fired straight up at a speed of 10 m/s, the total time to return to its starting position is about ?

When I use the quadratic formula, I get -8.97

How can the time be negative?

0=10t-4.9
0=t(10-4.9t)
t=0, t=10/4.9 sec are the solutions

In the quadratic equation:
(-10+-sqrt100)/-9.8 = same answers.

To calculate the total time it takes for a projectile to return to its starting position, you can use the equation of motion for vertical motion. In this case, the initial velocity is 10 m/s and the acceleration is due to gravity which is approximately -9.8 m/s^2 (taking into account its direction).

The equation of motion for vertical motion can be written as:

s = ut + (1/2)at^2

where:
s = vertical displacement (change in height)
u = initial velocity
a = acceleration due to gravity
t = time

For the projectile to return to its starting position, the vertical displacement should be zero.

0 = ut + (1/2)at^2

Rearranging the equation, we get a quadratic equation:

(1/2)at^2 + ut = 0

Substituting the values:
a = -9.8 m/s^2 (negative because it's in the opposite direction of initial velocity u)
u = 10 m/s

(1/2)(-9.8)t^2 + 10t = 0

Multiplying by 2 to eliminate the fraction:

-9.8t^2 + 20t = 0

Now, you can solve this quadratic equation to find the time it takes for the projectile to return to its starting position. You can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -9.8, b = 20, and c = 0.

Plugging these values into the quadratic formula, we have:

t = (-20 ± √(20^2 - 4(-9.8)(0))) / (2(-9.8))

Simplifying the equation further:

t = (-20 ± √(400)) / (-19.6)

Calculating the square root of 400:

t = (-20 ± 20) / (-19.6)

Now we have two possible solutions:

t1 = (-20 + 20) / (-19.6) = 0 / (-19.6) = 0
t2 = (-20 - 20) / (-19.6) = -40 / (-19.6) ≈ 2.04

When solving for time, we discard the negative value since time cannot be negative in this context. Therefore, the total time it takes for the projectile to return to its starting position is approximately 2.04 seconds.