When a ball is thrown in the air at 15mps from a height of 1 ft, how long does it take for the ball to hit the ground?

v=u-gt or 0=15-9.81t since final velocity is 0 and ball is going against gravity. Thus t=1.53 sec.

s=ut-.5gt^2=15-.5x9.81x(1.53)^2
s=22.95-11.48=121.47 which is the height the ball will go. From there it falls freely for a height of 11.47+1=12.47 ft.
s=0xt+.5x9.81xt^2 or t=(12.47x2/9.81)=1.59 secs.

PS: s=11.47 which has appeared as 121.47 at one place in 4th line.

since 1 ft = .3 m,

s(t) = .3 + 15t - 4.9t^2
it hits the ground when s=0, so

.3 + 15t - 4.9t^2 = 0
t = 3.08 sec

To determine the time it takes for the ball to hit the ground, we can use the equations of motion. In this case, we will use the equation:

h = v0 * t + (1/2) * g * t^2

Where:
h = initial height (1 ft)
v0 = initial velocity (15 m/s)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time

We want to find the time it takes for the ball to hit the ground, so we can set h equal to zero:

0 = v0 * t + (1/2) * g * t^2

Rearranging the equation, we get:

(1/2) * g * t^2 + v0 * t = 0

Now, we can solve this quadratic equation for t using the quadratic formula:

t = (-v0 ± √(v0^2 - 4*(1/2)*g*0)) / 2*(1/2)*g

Simplifying this equation further, we get:

t = (-v0 ± √(v0^2 - 2 * g * 0)) / g

Since we want the time it takes for the ball to hit the ground, the positive value of t is the solution we are interested in. Therefore, we use the positive sign before the square root:

t = (-v0 + √(v0^2 - 2 * g * 0)) / g

Substituting the given values:

t = (-15 + √(15^2 - 2 * 9.8 * 0)) / 9.8

Simplifying further:

t = (-15 + √(225 - 0)) / 9.8
t = (-15 + √225) / 9.8
t = (-15 + 15) / 9.8
t = 0 / 9.8
t = 0

Therefore, the time it takes for the ball to hit the ground is 0 seconds.