a model rocket is launched with an initial upward velocity of 39 m/s. the rocket’s height h (in meters)…

a model rocket is launched with an initial upward velocity of 39 m/s. the rocket’s height h (in meters) after t seconds is given by the following. \n( h = 39t - 5t^2 )\nfind all values of t for which the rocket’s height is 22 meters.\nround your answer(s) to the nearest hundredth. \n(if there is more than one answer, use the “or” button.)\n( t = ) seconds

a model rocket is launched with an initial upward velocity of 39 m/s. the rocket’s height h (in meters) after t seconds is given by the following. \n( h = 39t - 5t^2 )\nfind all values of t for which the rocket’s height is 22 meters.\nround your answer(s) to the nearest hundredth. \n(if there is more than one answer, use the “or” button.)\n( t = ) seconds

Answer

Explanation:

Step1: Set up the equation

We know that ( h = 39t - 5t^2 ) and we want to find ( t ) when ( h = 22 ). So we set up the equation: ( 22 = 39t - 5t^2 ) Rearranging this into standard quadratic form ( ax^2 + bx + c = 0 ), we get: ( 5t^2 - 39t + 22 = 0 ) Here, ( a = 5 ), ( b = -39 ), ( c = 22 ).

Step2: Use the quadratic formula

The quadratic formula is ( t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} ). First, calculate the discriminant ( D = b^2 - 4ac ). Substitute ( a = 5 ), ( b = -39 ), ( c = 22 ) into the discriminant formula: ( D=(-39)^2 - 4\times5\times22 ) ( D = 1521- 440 ) ( D = 1081 )

Then, find ( t ): ( t=\frac{39\pm\sqrt{1081}}{10} ) Calculate ( \sqrt{1081}\approx32.88 ) So we have two solutions: ( t_1=\frac{39 + 32.88}{10}=\frac{71.88}{10} = 7.188\approx7.19 ) ( t_2=\frac{39 - 32.88}{10}=\frac{6.12}{10}=0.612\approx0.61 )

Answer:

( t = 0.61 ) or ( t = 7.19 ) seconds