Enhanced EOC: Problem 2.20 A rock is tossed straight up from the ground with a s

Question

Enhanced EOC: Problem 2.20 A rock is tossed straight up from the ground with a speed of 22 m/s . When it returns, it falls into a hole 10 in deep. You may want to review (pages 51 - 54). For help with math skills, you may want to review: Quadratic Equations For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Time in the air for a tossed ball. Part A What is the rock?s velocity as it hits the bottom of the hole? Express your answer with the appropriate units. Part B How long is the rock in the air, from the instant it is released until it hits the bottom of the hole? Express your answer with the appropriate units.

Explanation / Answer

rock will go upwards and will have zero velocity at one point then it will start coming back to ground...

Let ground point to be origion and upward to be +y direction

so As V= u+ at

using this equation in y direction

0 = 22- 9.8t1

=> t1 = 2.24 sec,

so using symmetry total time to coming back on the ground

= 2*(2.24) sec

t1+t2 = 4.48 sec.

veloctity at the ground point when it come back on the ground will be 22 and in -Y direction as there is no energy losses in the middle.

So, let's solve equation for hole

V^2 = U^2 + 2as

=> V^2 = (22)^2 + 2 (-9.8)(-10)

=> V = 26.07 m/s ( ans to part 1)

now for time v= u+at

-26.07= -22+(-9.8)t

=>t = 0.415 sec

Hence total time in ait will be t1+t2+t3

4.48+.415 = 4.895 sec( ans for part 2)

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