Question

a ball thrown up is caught by the thrower after 4s. With what velocity was it thrown up? how high did it go? where was it after 3s?(g=9.8ms-square)​

Answer 1

Answer:

Explanation:

ATQ, height, s=19.6m

(i) Initial velocity, u=?

Now, when thrown vertically upwards,

Acceleration= −g=−9.8m/s

2

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

⇒v

2

=u

2

+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