Question

A stone is thrown vertically upward with an initial velocity of 50m/s. talking g=9.8m/s^2, find the maximum height reached by the stone, what is the net displacement and the total distance covered by the stone?

Answer 1

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.

Answer 2

Answer:

so,

given :

g=(-9.8m/s^2)<as stone is going against gravity>

u=50m/s

to find:

1)displacement

2)maximum height reached

3)total distance covered

1)

displacement is going to be 0m as the stone returns back to its place

2)

the formula for maximum height:

H=u^2/2g

H=(50)^2/2(9.8)    <DONT FORGET UNITS OR YOU WILL LOSE MARKS>                                

H=(2500)/19.6

H=127.551020m or H=127.55m

Therefore maximum height is equal to 127.55m

3) total distance will be double of maximum height reached as the stone returns back to it`s orignal place so it will be

=(127.55)2m

=255.1 m                   <DONT FORGET UNITS OR YOU WILL LOSE MARKS>  

thank you