The height after t seconds of an object projected upward with an initial velocity of 48 feet per second from a 210-foot tower can be modeled by h=−16t^2 + 48t +210. The height of a neighboring 50-foot tall building is modeled by the equation h=50. The time (t) when the object will be at the same height as the building is found to be t = –2 and t = 5. Which statement BEST describes the validity of these solutions? A. Neither solution is valid since time values cannot be squared. B. The solution t = – 2 is the only solution since 5 seconds is an unreasonable amount of time for the object to reach a height of 50 feet. C. The solution t = 5 is the only valid solution to this system since time cannot be negative. D. Both are valid solutions to this system since both values make the equation h=−16t^2+ 48t + 210 true.

Given:

The height h of an object after t seconds is

The height of a neighboring 50-foot tall building is modeled by the equation h=50.

The time (t) when the object will be at the same height as the building is found to be t = –2 and t = 5.

To find:

The statement which describes the validity of these solutions.

Solution:

We have,

Here, t is the time in seconds.

For t=-2,

For t=5,

So, the value of h is 50 at t=-2 and t=5.

We know that time is always positive so it cannot be negative value. It means t=-2 is not possible.

The solution t = 5 is the only valid solution to this system since time cannot be negative.

Therefore, the correct option is C.

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