a rocket is launched from atop a 105-foot cliff with an initial velocity of 156 ft/s.The height of the rocket above the ground at time or t is given by h=-16tsquared+156t+105. When will the rocket hit the ground after it is launched?

To determine when the rocket will hit the ground, we need to find the value of t when the height of the rocket, h, is equal to 0.

Given the equation for the height of the rocket:
h = -16t^2 + 156t + 105

We can set h to 0 and solve for t:
0 = -16t^2 + 156t + 105

This is a quadratic equation, which can be solved using various methods, such as factoring, completing the square, or using the quadratic formula. In this case, it may be easiest to use the quadratic formula:

The quadratic formula states that for an equation in the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)

Comparing our equation to the standard form ax^2 + bx + c = 0, we have:
a = -16, b = 156, c = 105

Substituting these values into the quadratic formula, we get:
t = (-156 ± √(156^2 - 4(-16)(105))) / (2(-16))

Simplifying further:
t = (-156 ± √(24336 + 6720)) / (-32)
t = (-156 ± √(31056)) / (-32)
t = (-156 ± 176) / (-32)

Now, we have two possible solutions for t:
1. t = (-156 + 176) / (-32)
2. t = (-156 - 176) / (-32)

Calculating each equation separately:
1. t = 20 / (-32) = -0.625
2. t = -332 / (-32) = 10.375

Therefore, the rocket will hit the ground after approximately 10.375 seconds from the time it is launched.

when is h= 0?

0=h(t)=-16t^2 + 156t+105

solve for t.