Explanation Given: A projectile is launched at an angle of θ hits an h-meters tower at a distance of d meters away. Proof: Let the launch point is at the origin, and r ( t ) be the position vector . Let the acceleration due to gravity be g. Then, v ( t ) = ∫ a ( t ) d t = 〈 0 , − g t 〉 + C 0 . ......... (1) Let the projectile was launched with speed of v 0 at an angle of θ . So, it gives v ( 0 ) = 〈 v 0 cos θ , v 0 sin θ 〉 . From (1), v ( 0 ) = 〈 0 , − g ⋅ 0 〉 + C 0 v ( 0 ) = 〈 0 , 0 〉 + C 0 v ( 0 ) = C 0 C 0 = 〈 v 0 cos θ , v 0 sin θ 〉 Therefore, the equation (1) becomes, v ( t ) = 〈 0 , − g t 〉 + 〈 v 0 cos θ , v 0 sin θ 〉 = 〈 v 0 cos θ , v 0 sin θ − g t 〉 Then, r ( t ) = ∫ 〈 v 0 cos θ , v 0 sin θ − g t 〉 r ( t ) = 〈 v 0 cos θ t , v 0 sin θ t − g 2 t 2 〉 + C 1 At t = 0 , r ( t ) = 〈 0 , 0 〉 . So, the equation r ( t ) = 〈 v 0 cos θ t , v 0 sin θ t − g 2 t 2 〉 + C 1 gives C 1 = 0

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Chapter 13.5, Problem 24E

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To show that when a projectile launched at an angle θ will hit the top of an tower of height h-meter located at a distance of d meters away if the initial speed is v0=g/2dsecθdtanθ−h.

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Chapter 13.5, Problem 23E

Chapter 13.5, Problem 25E

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Chapter 13 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

Ch. 13.1 - Prob. 1PQ Ch. 13.1 - Prob. 2PQ Ch. 13.1 - Prob. 3PQ Ch. 13.1 - Prob. 4PQ Ch. 13.1 - Prob. 5PQ Ch. 13.1 - Prob. 6PQ Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E

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To show that when a projectile launched at an angle θ will hit the top of an tower of height h-meter located at a distance of d meters away if the initial speed is v 0 = g / 2 d sec θ d tan θ − h .

Solution Summary: The author illustrates how a projectile launched at an angle of theta hits an h-meters tower at distance of d meters away.

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To show that when a projectile launched at an angle θ will hit the top of an tower of height h-meter located at a distance of d meters away if the initial speed is v0=g/2dsecθdtanθ−h.

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