Question

A ball thrown up with the velocity of 19.6ms-1 returns to thrower,s hand after​

Answer 1

Answer:

ATQ, height, s=19.6m

(i) Initial velocity, u=?

Now, when thrown vertically upwards,

Acceleration= −g=−9.8m/s2

Also, at end point, final velocity= 0m/s

⇒v2=u2+2as

So, (0)2=(u)2+2(−9.8)(19.6)

⇒u=19.6m/s

(ii) For going upward,

v=u+at

⇒0=19.6+(−9.8)(t)

⇒t=2s

Total time= upward+ downward

⇒2+2=4s

(iii) For downward motion,

u=0m/s

v=?

a=+9.8m/s

t=2s

⇒v=u+at

=19.6m/s