You hold two rocks, one in your left hand and one in your right hand. You toss t

Question

You hold two rocks, one in your left hand and one in your right hand. You toss the right-hand rock up and it reaches a maximum elevation that is about 1.6 times your height. At the same time that you toss up the first rock, you drop the rock in your left hand. You make sure to release the rocks from just about the same height (around the level of the top of your head).

5[Q]) Let the y y -axis represent the vertical direction, oriented so that y y increases in the upward direction. Which combination of statements is most accurate? [Enter your answer as a string of the letter options you believe are correct. For instance, if you think options A, D, and E are correct, then enter "ADE"]

A) The dropped rock lands in 2.07 times the time it took for the tossed rock to reach its maximum height.

B) The tossed rock lands in 2.04 times the time it took the dropped rock to land.

C) The tossed rock's speed just before it lands is 1.52 times the speed of the dropped rock just before it hit the ground.

D) The tossed rock's initial speed was equal to its speed as it descended back down past your head.

E) The dropped rock has traveled a quarter of your height after half the total time of its descent has passed.

Explanation / Answer

A)time by tossed rock to reach maximum height.

0=u2-2gh u=(2g*0.6h)1/2

v=u-gt=0

t1=(1.2h/g)1/2

time taken by dropped rock to reach ground=t2=(2h/g)1/2=1.29t1

B)time taken by tossed rock to land=2t1+[(1.2gh+2gh)1/2-(1.2gh)1/2]/g=2.884(h/g)1/2=2.039(h/g)1/2

C)tossed rock speed before it hitthe ground=v1=(1.2gh+2gh)1/2=(3.2gh)1/2

drop rock speed before hitting ground=v2=(2gh)1/2=

v1=1.265v2

D) thats true becoz v=u-gt

t=2u/g

v=u-g(2u/g)=-u

same velocity in opposite direction i.e. same speed

E)t=[(2h/g)1/2]/2

H=0.5gt2=0.5g(2H/g)/4=H/4

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