A stone is thrown vertically upward with an initial velocity of 49 m/s

1.	A stone is thrown vertically upward with an initial velocity of 49 m/s taking g=10 m/s² find the maximum height reached by the stone also find the time taken to reach that heightsteps explanation please
Answer» Answer: Initial VELOCITY u=40 Fianl velocity v=0 Height, s=? By THIRD equation of motion v 2 −u 2 =2gs 0−40 2 =−2×10×s s= 20 160 ⇒s=80m/s Toatl DISTANCE travelled by stone = upward distance + downwars distance =2×s=160m Total Diaplacement =0, Since the initial and final point is same.

A stone is thrown vertically upward with an initial velocity of 49 m/s taking g=10 m/s² find the maximum height reached by the stone also find the time taken to reach that heightsteps explanation please

Answer»

Answer:

Initial VELOCITY u=40

Fianl velocity v=0

Height, s=?

By THIRD equation of motion

−u

=2gs

0−40

=−2×10×s

160

⇒s=80m/s

Toatl DISTANCE travelled by stone = upward distance + downwars distance =2×s=160m

Total Diaplacement =0, Since the initial and final point is same.