Question

A ball thrown up is caught by the thrower $6$ s after start. The height to which the ball has risen is $(g=10m/s^2)$.

• A. $0$ m
• B. $30$ m
• C. $45$ m
• D. $90$ m

Answer 1

The correct option is C $45$ m

We know, $v=u-gt----\left(1\right)$

(For a body thrown against gravity)

The ball returns the same position after 6s of throwing.

Ascending time= descending time

t, for moving maximum height is 3s.

At maximum height, v=0

So, (1)=> 0= u-gt

=u- 10$\times$3

u= 30 m/s

Now, s= ut- $\frac{1}{2}$g$t^2$

s= (30$\times$3)- 0.5$\times$10$\times$$3^2$

90-45

= 45m