8. janet throws a ball into water below from the top of a cliff that is 175 m high. the height of the ball…

8. janet throws a ball into water below from the top of a cliff that is 175 m high. the height of the ball above the water after it is thrown is modelled by the function h = - 5t² - 10t + 175, where h is the height in metres and t is the time in seconds. do not use quadratic formula. t - 6,c - 2 a) when will the ball hit the water below? b) when will the ball reach the ledge that is 100 m above the water? 9. at a blue jays game, a baseball player hits a ball into the air. the height of the ball at a given time is modelled by the function h=-4.9t² + 22t + 1.2, where t is the time, in seconds, and h is the height of the ball, in metres. t - 5,c - 2 a) what is the height of the ball after 2 seconds? b) when will the ball hit the ground?

8. janet throws a ball into water below from the top of a cliff that is 175 m high. the height of the ball above the water after it is thrown is modelled by the function h = - 5t² - 10t + 175, where h is the height in metres and t is the time in seconds. do not use quadratic formula. t - 6,c - 2 a) when will the ball hit the water below? b) when will the ball reach the ledge that is 100 m above the water? 9. at a blue jays game, a baseball player hits a ball into the air. the height of the ball at a given time is modelled by the function h=-4.9t² + 22t + 1.2, where t is the time, in seconds, and h is the height of the ball, in metres. t - 5,c - 2 a) what is the height of the ball after 2 seconds? b) when will the ball hit the ground?

Answer

Explanation:

Step1: Solve 8a

Set (h = 0) in (h=-5t^{2}-10t + 175). Factor out - 5: (-5(t^{2}+2t - 35)=0). Then (t^{2}+2t - 35 = 0). Factor the quadratic: ((t + 7)(t - 5)=0). Set each factor equal to zero: (t+7 = 0) gives (t=-7) (rejected as time can't be negative), (t - 5=0) gives (t = 5).

Step2: Solve 8b

Set (h = 100) in (h=-5t^{2}-10t + 175). We get (-5t^{2}-10t + 175=100). Rearrange to (5t^{2}+10t - 75 = 0). Divide by 5: (t^{2}+2t - 15=0). Factor: ((t + 5)(t - 3)=0). Set each factor equal to zero: (t+5 = 0) gives (t=-5) (rejected), (t - 3=0) gives (t = 3).

Step3: Solve 9a

Substitute (t = 2) into (h=-4.9t^{2}+22t + 1.2). (h=-4.9\times2^{2}+22\times2 + 1.2=-4.9\times4+44 + 1.2=-19.6+44 + 1.2=25.6).

Step4: Solve 9b

Set (h = 0) in (h=-4.9t^{2}+22t + 1.2). Factor out - 0.1: (-0.1(49t^{2}-220t - 12)=0), so (49t^{2}-220t - 12 = 0). Using factoring or other non - quadratic formula methods (by inspection or grouping), we find the roots. We can also use the quadratic - like factoring approach. Multiply (a = 49) and (c=-12) to get (-588). We need two numbers that multiply to (-588) and add up to (-220). By trial and error or other methods, we find the roots. Another way is to rewrite the equation as (49t^{2}-224t + 4t-12 = 0), (7t(7t - 32)+ \frac{4}{7}(7t - 32)=0), ((7t - 32)(7t+\frac{4}{7}) = 0). (7t-32 = 0) gives (t=\frac{32}{7}\approx4.57), (7t+\frac{4}{7}=0) gives (t=-\frac{4}{49}) (rejected as time can't be negative).

Answer:

8a. (t = 5) seconds 8b. (t = 3) seconds 9a. (h = 25.6) metres 9b. (t=\frac{32}{7}\approx4.57) seconds