Assume the acceleration of the object is a(t)= -9.8 meters per second. (Neglect air resistance.) With what initial veloc

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: Assume The Acceleration Of The Object Is A T 9 8 Meters Per Second Neglect Air Resistance With What Initial Veloc 1 (27.57 KiB) Viewed 26 times

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: Assume The Acceleration Of The Object Is A T 9 8 Meters Per Second Neglect Air Resistance With What Initial Veloc 3 (24.93 KiB) Viewed 26 times

Assume the acceleration of the object is a(t)= -9.8 meters per second. (Neglect air resistance.) With what initial velocity must an object be thrown upward (from a height of 2 meters) to reach a maximum height of 680 meters? Step 1 We want to find the initial velocity, v(0) of an object thrown upward such that the maximum of the height function () is 680. We also assume the acceleration due to gravity is att) -9.8 meters per second. First, we must find an equation for the velocity and height functions. Recall that the derivative of the velocity function is acceleration, or v(t)= a(t). This also means that it is the antiderivative of the constant function a(t). Find v(t). v(t)= a(t) dt -9.8 d -98 )t+C The initial velocity is the velocity at time t 0. Find the initial velocity by substituting 0 for t in the equation above. v(0) = (0) - 9.8✔ Step 2 We have found that the velocity of the object thrown upward has the general solution v(t)-9.8t+ C, and the initial velocity is v(0) - C Substituting v(0) for the constant C to find the particular solution for the velocity gives us the following. v(t)-9.8t + (0) Now recall that the derivative of the height function is the velocity, or s'tt) vt). This also means that s(t) is the antiderivative of v(t). Find (f) - v(t) of [1-981 (-9.8t v(0)) at

Step 2 We have found that the velocity of the object thrown upward has the general solution () -9.80 C, and the initial velocity is (0) C Substituting (0) for the constant C to find the particular solution for the velocity gives us the following v(e)-9.8+ (0) Now recall that the derivative of the height function is the velocity, or st) (). This also means that sr) is the antiderivative of (1) Find s (2) - Ives a . (-9.8t (0)) ot • f(-x+ (✓ 14² cose + ₁ Step 3 We now know that the general solution of the height function is s(t)-4.9²(0)+c. We also know that the inmal height of the otject is 2 meters. In other words, Use the initial height of the object to solve for the constant C₂ (0) - ✓ +v(0)(0) + C₁ 2-00+C₁ =C₁ Step 4 Remember that we are looking for the value of the initial velocity, v(0), if the maximum height is 680 meters, or s(t)- 680 is a maximum for some t If a(t) is a maximum, then for this value of rt the value of the velocity function is (t)-✓ Substitute this value for v(n) and selve for t (t)-9.8t+ (0) ARCHER

Stap 4 Remember that we are looking for the value of the initial velocity, v(0), if the maximum height is 650 meters, or (t)- 680 is a maximum for some t 0 Substitute this value for v(t) and solve for t If sit) is a maximum, then for this value of t the value of the velocity function is v(t)-✔ vt) -9.8+ (0) 0-9.at (0) 9.8✔ 10. 9.8 :-(0) DNE Step 5 (0) 9.8 We have found that the particular solution of the height function is s(t)-4.91² v(0)t +2 and s(t)- 680 is the maximum height when t- Substitute this value of it in the height function to solve for v(0), rounding the final result to one decimal place. (Enter a positive value for your final answer) 1.91²+ (0)2 M(C) 680- 678- 678- -4.9 (૦૭(-- (0)² (0) 49 9.8 9.8² Subinit Skip (v cannot come back . (0) X²)-(²) 9.8² x=v(0) Therefore, from an initial height of 2 m, the initial velocity (in m/sec) that the object must be thrown upward to reach a maximum height of 680 meters is as follows. (Round your answer to one decimal place.) DNE x m/sec +2