Question

A pebble is thrown vertically upward from a bridge with an initial velocity of 4.9 m/s .It strikes the water​

Answer 1

A pebble is thrown vertically upward with an initial velocity of 4.9 m/s from a bridge . After 2 s it falls down in the water. What is the height of the bridge?

s=ut+12at2, where, s,u,t and a are the displacement, the initial velocity, the time taken and the acceleration.

It is given that u=4.9 m/s and t=2 s.

Further, g=−9.8 m/s2, the negative sign indicating that it is in the downward direction.

⇒s=4.9×2+12×(−9.8)×22=4.9(2−4)=−9.8 m.

Note that the displacement is negative indicating that the stone is below the bridge.

⇒ The height of the bridge is 9.8 m.

Answer 2

u=4.9m/s²

t=2s

a=9.8m/s²

s=ut+1/2gt²

s=4.9X2+1/2X(9.8)X(2)²

s=9.8+1/2X9.8X4

s=9.8+9.8X2 {we can divide 4 by 2}

s=9.8-19.6 {height kitni hai h toh upar se jakar vo niche gir Rahi hai toh - aayega}

s=9.8m (is question may 4.9 ki sign kya hai vo depend karta hai + raha toh answer+ve ,-ve raha toh answer -ve)

v=u+at

v=4.9-9.8X2

v=4.9-19.6

v=14.7m/s (jo s final answer nikla tha usme Jo bataya bracket mein same)