Can anyone help me with this question The answer for a is 5

Can anyone help me with this question?

The answer for a) is 5 sec and the answer for b) is (3, infinity) but I don\'t know how to deriive those answers.

The functions(t)--49:2 +49t + 58.8 meters, of an object launched with a velocity of 4.9 m/s froma height of 58.8 m. a) How long will it take the object to hit the ground? b) Find the interval on which the height of the object is greater 21· gives the height S, in than 49 m.

Solution

S(t) = -4.9t^2 + 4.9t + 58.8

a) Time to hit the groudn can be derived by plugging S(t) = - 58.8m and solve for t

-4.9t^2 + 4.9t + 58.8 = -58.8

-4.9t^2 + 4.9t + 117.6 =0

Use quadratiec formula to solve the above equation :

t = ( -4.9 + /- sqrt( 4.9^2 +4*4.9*117.6) / -2*4.9

t = 5.42 ,-4.4

Neglect the -ve value of t

So, t= 5.42 sec or 5 sec (rounded of to 1 nearest whole no.)

b) S(t) > 49

-4.9t^2 + 4.9t + 58.8 > 49

-4.9t^2 + 4.9t +9.8>0

Solve the inequality :

(-49t^2 + 49 +98)/10 >0

-49t^2 + 49 +98 <0

49( t+1)(t -2) <0

So, solution : -1 < t< 2

Neglect -ve time range

So, solution ( 0, 2)

I am getting solutuon as ( 0 sec ---2 seconds when height is greater than 49 mt