Question

a stone thrown upward returns back to the thrower 6s . The maximum height attained by the stone is. Take G=10m/s^-2

Answer 1

Answer:

Total time of flight(t)=6s

So time to reach the maximum height =3s

if u= projected velocity then:

0=u−g

2

t

⇒u=

2

gt

=

2

60

=30 m/s

Applying equation v

2

=u

2

−2gh(for half journey)

0=900−20h

⇒h=45m

When the ball is at height 5m from highest point, it's height from ground =45−5=40m

Applying equation: s=ut+

2

1

at

2

(at h=40m)

⇒40=30t−

2

1

gt

2

⇒40=30t−5t

2

⇒t

2

−6t+8=0

As one root of the equation is given as: t

1

=2s

By product of roots t

1

t

2

=8

⇒t

2

=4s. (when the ball reaches the same height again)