Motion in A Straight Line Test

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An inquisitive student, determined to test the law of gravity of for himself. walks to the top of a building of $$150$$ floors, with every floor of height $$4\ m$$ having a stopwatch in his hand (the $$1^{st}$$ floor is at a height of $$4\ m$$ from the ground level). From there he jumps off with negligible speed and has start rolling freely. A rocketeer arrive at the scene $$5\ s$$ later and dives off from the top of the building to save the student. The rocketeer leaves the roof with an initial downward speed $$v_0$$. In order to catch the student and to prevent injury to him, the rocketeer should catch the student at a sufficiently great height above ground so that the rocketeer and the student slow down and arrive at the ground with zero velocity. The upward acceleration that accomplishes this is provide by rockteer's jet pack, which he terms on just as he catches the student, before the rocketeer is in free fall. To prevent any discomfort to the atudent, the magnitude of the acceleration of the rocketeer and the student as they move downward together should not exceed $$5\ g$$.

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Motion in A Str...

Question 1

45m

90m

72m

135m

Question 2

h/9 m from the ground

7h/9 m from the ground

8h/9 from the ground

17h/18 m from the ground

Question 3

$$12 \ s$$

$$8 \ s$$

$$5 \ s$$

$$16 \ s$$

Question 4

$$a(1-b^{2t})$$

$$\dfrac{a}{b}(1-e^{-bt})$$

$$abe^{-t}$$

$$ab^2(1-t)$$

Question 5

$$zero$$

$$40\sqrt { 2 } $$ km/h north-east

$$40\sqrt { 2 } $$ km/h south-east

$$None$$

Question 6

Towards east

towards west

towards north

towards south

Question 7

$$5g\ ms^{-2}$$

$$2g\ ms^{-2}$$

$$4g\ ms^{-2}$$

$$Cannot\ be\ determined$$

Question 8

$$4.8$$ sec

$$8$$ sec

$$10$$ sec

$$12$$ sec

Question 9

The particle has a constant acceleration

The particle has never turned around

The average speed in the interval $$0$$ to $$10$$ s in the same as the average speed in the interval $$10$$ s to $$20$$ s

Both (a) and (c) are correct

Question 10

$$ \dfrac{v^{2}}{a} $$

$$ \dfrac{v^{2}}{2a} $$

$$ \dfrac{2v^{2}}{a} $$

$$ \dfrac{v^{2}}{4a} $$

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