Afirework is shot into the air from an initial height of 6 feet. the height in feet of the firework above the ground is given by s(t)=-15t^2+52t+6, where t is time, in seconds, and t≥0, at what time will the firework be 45 feet off the ground?

Given function is

where t is the time in seconds and s(t) is the height of the firework in t seconds.

Given that height of the firework is 45 feet. Now we have to find when it happens that is find value of t when s(t)=45

Apply quadratic formula

or

or

or

Hence final answer is 2.3693 seconds and 1.09737 seconds.

Two answers means height of 45 feet will be attained two times.

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

25.2

step-by-step explanation:

2+4x6-8/10= 25.2