6 -16 pointsTGInterAlghHS 5.3.02. EXAMPLE 2 If a toy rocket is shot straight up

Question

6 -16 pointsTGInterAlghHS 5.3.02. EXAMPLE 2 If a toy rocket is shot straight up with an initial velocity of 128 feet per second, its height, in feet, t seconds after being launched is given by the function below. h(t)--162+128 Find the height of the rocket (a) 2 seconds after being launched (b) 7.9 seconds after being launched. Strategy We will find n(2) and (7.9). The notation hK2) represents the height of the rocket 2 seconds after being launched and (7.9) represents the height of the rocket 7.9 seconds after being launched. Why Solution (a) To find the height of the rocket 2 seconds after being leunched, we need to evaluate the function at t 2. That is, we need to find h(2) h(c)-16(02 128 This is the given function h(2) =-16(2)2 + 128(2) Substitute 2 for each t. (The rput is 2.) --16(4)Evaluate the right-hand side. +256 The output is 192 We have found that h(2) -192. Thus, 2 seconds after it is launched, the height of the rocket is 192 feet. (b) To find the height of the rocket 7.9 seconds after it is launched, we find h(7.9) as follows. h(t)--16(C)2 + 128r h(7.9)--16(7.9)2+ 128(7.9) This is the given function. Substitute 7.9 for each t. (The input is 7.9.) --161,011.2 Evaluate the right-hand side. -998.56+ -12.64 The output is 12.64 At 7.9 seconds, the height of the rocket is 12.64 feet. It has almost fallien back to Earth. Self Check 3 Find the height of the rocket 4 seconds after it is launched ft http

Explanation / Answer

h(t) = -16t^2 + 128 t

a) h(2) = - 16(2)^2 + 128 (2)

= - 16(4) + 256

= - 64 + 256

= 192

b) h(7.9)

h(7.9) = -16(7.9)^2 + 128(7.9)

= -16 ( 62.41 ) + 1011.2

= -998.56 + 1011.2

= 12.64

c) height of the rocket 4 seconds after

h(4) = - 16(4)^2 + 128 (4)

= -16 (16) + 512

= - 256 + 512

= 256

height after 4 seconds = 256 feet

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