The height, in meters, of a projectile can be modeled by h= -4.9t^2 + vt + s where t is the time (in seconds) the object has been in the air, v is the initial velocity (in meters per seconds) and s is the initial height (in meters). A soccer ball is kicked upward from the ground and flies through the air with an initial velocity of 4.9 meters per second. Approximately, after how many seconds does it land?
so your height formula is
h = -4.9t^2 + 4.9t + 0 , assuming kicked from ground
when it lands, h = 0
-4.9t^2 + 4.9t = 0
-4.9t(t - 1) = 0
t = 0 ----> at the moment the ball is kicked
or
t = 1, it lands 1 second later
To determine the time it takes for the soccer ball to land, we need to find the value of t when the height (h) becomes zero. The equation given to model the height is h= -4.9t^2 + vt + s, where h represents the height in meters, t represents the time in seconds, v represents the initial velocity in meters per second, and s represents the initial height in meters.
In this case, the ball is kicked upward from the ground, so the initial height (s) would be zero. The equation simplifies to h= -4.9t^2 + vt. We can set h equal to zero to find the time at which the ball lands:
0 = -4.9t^2 + vt
Now, we can rearrange the equation to make it easier to solve for t:
4.9t^2 = vt
Dividing both sides of the equation by t, we get:
4.9t = v
Substituting the value of v (initial velocity of 4.9 m/s) into the equation, we have:
4.9t = 4.9
Now, divide both sides of the equation by 4.9:
t = 1
Therefore, the soccer ball will land after approximately 1 second.