A rocket is launched from atop a 65-foot cliff with an initial velocity of 113 ft/s. The height of the rocket above the ground at time t is given by. When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second.

7.6 seconds

Step-by-step explanation:

To solve this problem we can use the following equation:

S = So + 113*t + at^2/2

Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time.

In this problem, we have that S = 0, So = 65, Vo = 113 and a = -32 ft/s2 (The acceleration of gravity)

So we have that:

0 = 65 + 113t - 16t2

16t2 - 113t - 65 = 0

Using Bhaskara's formula, we have:

Delta = 113^2 + 4*65*16 = 16929

sqrt(Delta) = 130.11

t = (113 + 130.11) / (2*16) = 7.597 seconds

Rounding to nearest tenth, we have t = 7.6 seconds

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step-by-step explanation:

step-by-step explanation:

two pairs of adjacent sides are of equal length. this implies that one diagonal divides the kite into congruent triangles, and so the angles between the two pairs of equal sides are equal in measure. it also implies that the diagonals are perpendicular.