Question

A stone is thrown up from the surface of the earth with an initial speed of the earth with an initial speed 5m\ sec. The stone comes to rest at a height of cas sume g= 10m\sec

Answer 1

Answer:

s=1.25 m

Explanation:

u=initial velocity= 5m/s

v=final velocity=0m/s(as object comes at rest)

a=acceleration due to gravity =-g =

-10m/s²

s=displacement of the object during half motion (from bottom to top)

v²-u²=2as

0²-5²=2(-10)*s

-25=-20*s

s= 1.25m

therefore the object attains a height of 1.25 m before coming at rest :)

Answer 2

Answer:

1.25m

Explanation:

Maximum height attained by a body thrown up with initial velocity u is given by

${u}^{2} \div 2g$

so, u = 5m/sec

g = 10(approx)

Hence, H (max)= 5*5÷2*10= 1.25 (approx)

For exact values you can replace g by9.8m/sec^2