At the Atlanta United game, a ball flew into the stands. A fan launches the ball at 19.6 feet per second (ft/s) from a 60 ft tall bleacher. The equation for the soccer ball's height, s, at time, t, seconds after the launch is s(t) = −4.9t2 + 19.6t +60. When does the ball hit the ground?

6.03s

Step-by-step explanation:

Given the equation for the soccer ball's height, s, at time, t, seconds after the launch expressed as

s(t) = −4.9t2 + 19.6t +60

The ball hits the ground when s(t) = 0

Substitute s(t) = 0 into the expression

0= −4.9t2 + 19.6t +60

Multiply through by -10

0 = 49t²-196t-600

49t²-196t-600 = 0

Factorize

t = 196±√196²+4(600)(49)/2(49)

t = 196±√38416+117600/98

t = 196±√156016/98

t =196±394.99/98

t = 196+394.99/98

t = 590.99/98

t = 6.03secs

Hence the ball bit the ground after 6.03seconds later

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

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