Question

a stone thrown upward attaims a maximum height of 39.2m. Find the velocity with which it was thrown

Answer 1

Answer:

The velocity with which it was thrown is 27.718m/s

Explanation:

we know that V²=U²+2aS

where V is the final velocity, U is the initial velocity, a is the acceleration, S is the distance covered by stone.

V=0 since its final velocity is zero we are left with only U, a and S

Note:Remember that here acceleration is negative.

0=U²+2*(-9.8)*39.2

U²=-2*-(9.8)*39.2

U²=768.32

Rooting both sidea we get

U =27.71m/s.