A baseball player releases a baseball at the height of 7 feet above the ground with an initial vertical velocity of 54 feet per second. How long will it take the ball to reach the ground?

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To find the time it takes for the ball to reach the ground, you can use the equation of motion:

h(t) = h0 + v0t - (1/2)gt^2

Where:
h(t) is the height of the ball at time t
h0 is the initial height of the ball
v0 is the initial vertical velocity of the ball
g is the acceleration due to gravity

In this case:
h0 = 7 feet (initial height)
v0 = 54 feet per second (initial vertical velocity)
g = -32.2 feet per second squared (acceleration due to gravity, negative because it acts downwards)

We want to find the time it takes for the ball to reach the ground, which means h(t) will be 0 (since the height of the ground is 0).

0 = 7 + 54t - (1/2)(-32.2)t^2

Simplifying this equation, we get a quadratic equation:

16.1t^2 - 54t - 7 = 0

Now, you can solve this quadratic equation to find the time it takes for the ball to reach the ground. You can use the quadratic formula or factoring to solve it. In this case, factoring might be more challenging, so let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Considering the equation, a = 16.1, b = -54, and c = -7, we can plug in these values into the quadratic formula to find the solutions for t.

Vf^2 = Vo^2 + 2gd,

d = (Vf^2 - Vo^2) / 2g,
d = 7 + (0 - (54)^2) / -64 = 52.56 Ft.
= Distance above ground.

Vf = Vo + gt,
t(up) = (Vf - Vo) / g,
t(up) = (0 - 54) / -32 = 1.69s.

d = Vo*t + 0.5gt^2 = 52.56 Ft.
0 + 0.5*32t = 52.56,
16t = 52.56,
t(down) = 3.29s.

T = t(up) + t(down) = 1.69 + 3.29 = 4.98s = Time to reach ground.

Correction:

0 + 0.5*32t^2 = 52.56,
16t^2 = 52.56,
t^2 = 3.29,
t(down) = 1.81s.

T = t(up) + t(down) = 1.69 + 1.81 = 3.50s. = Time to reach ground.