21. A rocket, initially at rest, is fired vertically with an upward acceleration

Question

21. A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s. At an altitude of 0.50 km, the engine of the rocket cuts off. What is the maximum altitude it achieves? a. 1.9 km b. 1.3 km c. 1.6 km d. 1.0 km e. 2.1 km An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object? 23. a. 35 m/s b. 25 m/s c. 30 m/s d. 21 m/s e. 17 ms An object is thrown downward with an initial (t-0) speed of 10 m/s from a height of 60 m above the ground. At the same instant (t-0), a second object is propelled vertically upward from ground level with a speed of 40 m/s. At what height above the ground will the two objects pass each other? a. 53 m b. 41 m c. 57 m d. 46 m e. 37 mm 25. A rock is thrown downward from an unknown height above the ground with an initial speed of 10 m/s. It strikes the ground 3.0 s later. Determine the initial height of the rock above the ground. 27. a. 44 m b. 14 m c. 74 m d. 30 m e. 60 m An object is thrown vertically upward such that it has a speed of 25 m/s when it reaches two thirds of its maximum height above the launch point. Determine this 29. maximum height. a. 64 m b. 48 m c. 32 m d· 96m e. 75 m

Explanation / Answer

21)  By s = ut + 1/2at^2
=>0.50 x 1000 = 0 + 1/2 x 10 x t^2
=>t = 100 = 10 sec
=>By v = u + at
=>v = 0 + 10 x 10 = 100 m/s
=>By v^2 = u^2 - 2gh
=>0 = (100)^2 - 2 x 9.8 x h
=>h = 10000/19.6 = 510.20 m
=>Total height = 500 + 510.20 = 1010.20 m = 1.01 Km

23)  V = original velocity

1/2 . m . V^2 = original KE = Final Height energy = m . g . H

g . H = V^2 / 2

KE at 3/4 H = 1/2 . m . 18^2 = energy needed to get up the remaining 1/4 H = 1/4 . m . g . H

1/2 . 324 = 1/4 . V^2 / 2

V^2 = 1225

V = 35 m/s

27) d = vi * t + .5 * a * t^2
d = 10 * 3 + .5 * 9.8 * 3^2
d = 22.25 + 30.625
d = 74 m

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