A toy rocket is launched straight up into the air with an initial velocity of 60 ft/s from a table 3 ft above the ground. If acceleration due to gravity is –16 ft/s2, approximately how many seconds after the launch will the toy rocket reach the ground?

h = -16t^2 + 60t + 3

at ground level , h = 0

16t^2 - 60t - 3 = 0
t = (60 ± √3792)/32
t = 3.799.. or t = a negative

it will take appr 3.8 seconds to hit the ground

To find out how many seconds the toy rocket will take to reach the ground, we can use the kinematic equation for displacement:

s = s0 + v0t + (1/2)at^2

where:
s is the displacement (distance traveled),
s0 is the initial position,
v0 is the initial velocity,
a is the acceleration, and
t is the time.

In this case, the initial position s0 is 3 ft, and the acceleration a is -16 ft/s^2 (negative because it is acting against the direction of motion).

The final position s is 0 because the toy rocket will reach the ground.

So, the equation becomes:

0 = 3 + 60t + (1/2)(-16)t^2

Let's solve this equation for t.

First, simplify the equation:

0 = 3 + 60t - 8t^2

Next, rearrange the equation to get it in the standard form for a quadratic equation:

8t^2 - 60t - 3 = 0

Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = 8, b = -60, and c = -3.

Plugging these values into the quadratic formula:

t = (-(-60) ± √((-60)^2 - 4(8)(-3))) / (2(8))

Simplifying:

t = (60 ± √(3600 + 96)) / 16
t = (60 ± √(3696)) / 16

Let's evaluate this expression:

t = (60 ± √(3696)) / 16
t ≈ (60 ± 60.8) / 16

Now, we have two possible values for t: (60 + 60.8) / 16 and (60 - 60.8) / 16.

t1 ≈ (60 + 60.8) / 16 ≈ 7.55
t2 ≈ (60 - 60.8) / 16 ≈ -0.05

Since time cannot be negative, we discard the negative value.

Therefore, approximately 7.55 seconds after the launch, the toy rocket will reach the ground.

To solve this problem, we can use the kinematic equation that relates displacement, initial velocity, time, and acceleration:

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

Where:
s = displacement (distance traveled)
u = initial velocity (60 ft/s)
t = time (unknown)
a = acceleration due to gravity (-16 ft/s^2)

Since the toy rocket starts 3 ft above the ground and eventually reaches the ground, the displacement s is equal to -3 ft (negative because the displacement is in the opposite direction to the positive y-axis).

Plugging these values into the equation, we have:

-3 = (60)t + (1/2)(-16)t^2

Simplifying the equation and rearranging it to the standard quadratic form (ax^2 + bx + c = 0), we get:

-16t^2 + 60t - 3 = 0

Now, we can use the quadratic formula to solve for t:

t = (-b ± √(b^2 - 4ac)) / 2a

Plugging in the values for a, b, and c, we have:

t = (-(60) ± √((60)^2 - 4(-16)(-3))) / 2(-16)

Simplifying this equation, we get:

t = (-60 ± √(3600 - 192)) / (-32)

t = (-60 ± √(3408)) / (-32)

t = (-60 ± 58.38) / (-32)

Using the positive value from the ±, we have:

t = (-60 + 58.38) / (-32)

t = -1.38 / -32

t ≈ 0.043 seconds

Therefore, approximately 0.043 seconds after launch, the toy rocket will reach the ground.