A stone is thrown vertically upward with a speed of 27.0m/s . How fast is it moving when it is at a height of 13.0m ? How much time is required to reach this heigh

Question

A stone is thrown vertically upward with a speed of 27.0m/s .  How fast is it moving when it is at a height of 13.0m ?  How much time is required to reach this heigh

anonymous · Answer

u can use third equation of motion

anonymous · Answer

$$v ^{2} + u ^{2} = 2gs$$

anonymous · Answer

$$v ^{2} = 27^{2} + 2*10*13$$

anonymous · Answer

$$v=31.44 m/s$$

anonymous · Answer

Now u can use first equation to calculate time. i.e. v = u + at
31.44= 27 + 10t
10t = 4.44
t= .444s

anonymous · Answer

plz vote 4 me if u are satisfied

agent0smith · Answer

@VK260 you made a mistake, gravity should be a negative number not positive

anonymous · Answer

The kinetic energy of the stone is proportional to its velocity squared (v^2). The loss of kinetic energy is due to the work done on the stone by gravity, proportional to 2gs, where g is gravitational acceleration and s is the distance. The equation needed is v^2=v0^2-2gs.  (^2 means square).
v^2=(27)^2-2(9.8)(13)= 4742 (m/s)^2 and v=21.8 m/s, which is less than the initial velocity, as it should be. [All my units are mks.]

Gravity is decelerating the stone at 9.8m/s^2 so its velocity as a function of time is v=v0-(9.8)t, which we can solve to get t=(21.8-27.0)/(-9.8) =
0.53s. We can also see that it stops at t=v0/9.8=2.76s.