Question

A stone thrown vertically upward passes a certain

point P at the end of 2 seconds and 8 seconds re-

spectively. Find the maximum height reached by the

stone.​

Answer 1

Explanation:

A stone thrown vertically passes a certain point P at the end of 2 seconds and 8 seconds respectively. Find the maximum height reached by the stone .(Takeg=10ms

−2

)

ANSWER

Let initial velocity u and point 'P' is at height h.

At the end of 2 seconds,

Using s=ut+

2

1

at

2

⇒h=u×2−

2

1

g×2

2

⇒h=2u−20 .......1

At the end of 8 seconds,

⇒h=u×8−

2

1

g×8

2

⇒h=8u−320 ....2

Solving equations 1 and 2, ⇒u=50m/s,

At maximum height , velocity v=0,

Using v

2

=u

2

+2as ⇒0=50

2

−2×10×s ⇒s=125m,

∴ Maximum height =125m