Question

22. For the following equation of the path of a rocket h ( t ) = − 9.8 t 2 + 78.4 t + 196, where height, h, is measured in meters and time, t, is measured in seconds, what is the height from which it was launched (t=0)?

23.For the following equation of the path of a rocket h ( t ) = − 9.8 t 2 + 78.4 t + 196, where height, h, is measured in meters and time, t, is measured in seconds, how long after launch does it hit the ground?

24.For the following equation of the path of a rocket h ( t ) = − 9.8 t 2 + 78.4 t + 196, where height, h, is measured in meters and time, t, is measured in seconds, how long after launch does it reach its maximum height?

25.For the following equation of the path of a rocket h ( t ) = − 9.8 t 2 + 78.4 t + 196, where height, h, is measured in meters and time, t, is measured in seconds, what is the maximum height?

26.For the following equation of the path of a rocket h ( t ) = − 9.8 t 2 + 78.4 t + 196, where height, h, is measured in meters and time, t, is measured in seconds, how long after launch does it reach 196 meters?

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Answer 1

Equation of the path of rocket hit -9.842 +78.4€ + 196 _22) hught when tzo is h(o)= -9.8(0)²+ 78.4(0) + 196 = 0+0+ 196 =196 = 196 mis 23) The rocket hits ground when h(t)=0 oz 9.8t ² + 78.44 +196 9.862_78.45-196=0 telo 31 t=-2 t cannot be negative te lo secs For maximum height hiltzo hilt = -9:8(26) +78.4(1) o= -19.6t + 78.4 19.6t = 78.4 t = 4 secs Maximum height is at tau secs

(25) Maximum height is at t=4secs h (4) = -9.8(4) ²+ 78.4(4) + 196 = -9.8 (16) + 78-4 (41 +196 - 156.8+ 313.6 + 196 = 352.8 mts Maximum hught 352-8 mts (26) Given height of rocket - 196 units hlt) = 196 -9.8t²+78.46 +196 - 196 - 9.842 = -78.4t toi -9.88 = -78.4 t=8 sec After te 8 secs it reaches 196 mts