1 ) A rock is tossed straight up from the ground with a speed of 25 m/s . When i

Question

1 ) A rock is tossed straight up from the ground with a speed of 25  m/s . When it returns, it falls into a hole 10 m deep.
A) What is the rock's velocity as it hits the bottom of the hole?
   B) How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?


2) A particle moving along the x-axis has its position described by the function x =( 5.00 t3 1.00t+ 5.00 )m, where t is in s. At t= 2.00, what are the particle's
     a) position
     b) velocity
     c) acceleration?

Explanation / Answer

1. initial speed=v=25 m/s
as we know, the speed at the ground level will be 25 m/s (applying conservation of energy, change in potential energy in zero, hence kinetic energy will remain the same an dhence speed will remain the same)

now while falling into the hole, the accleration=9.8 m/s^2

hence velocity when it hits bottom, let be v2.

then v2^2-25^2=2*9.8*10

v2=28.65 m/s

b)time taken to reach the ground level =2*initial speed/g=2*25/9.8=5.102 seconds

time taken to travel 10 m in the hole, let be, t
then 28.65=25+9.8*t

t=0.372 seconds

so total time travelled=5.102+0.372=5.474 seconds

Q2. position:
x=5*t^3-t+5
velocity=dx/dt=15*t^2-1

acceleration=30*t

at t=2,

position=43 m
velocity=59 m/s
acceleration=60 m/s^2

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