Question

An object is dropped from the top of a tower. It
travels a distance 5 m in the first second and 25 m
in the last second. The speed with which object
strikes the ground is (g = 10 m/s2)
O 30 m/s
O 50 m/s
O 40 m/s
O 20 m/s​

Answer 1

Given  :  An object is dropped from the top of a tower.

It travels a distance 5 m in the first second and 25 m in the last second.

To Find :  The speed with which object strikes the ground is (g = 10 m/s2)

O 30 m/s

O 50 m/s

O 40 m/s

O 20 m/s​

Solution:

object is dropped from the top of a tower

u = 0

Distance travelled in 1st sec

S = ut + (1/2)at²

t = 1 sec  a = g = 10 m/s²

=> S = 0 + (1/2) * 10 * 1²  = 5 m

Verified

Distance travelled in last sec = 25 m

Let sat velocity  in beginning of  last sec  =  v

S = 25

t = 1

25 = v(1) + (1/2) (10)(1)²

=> 25 = v  +  5

=> v = 20

V = U + at

Velocity while striking ground =  20 + (10) (1)

= 20 + 10

= 30  m/s

The speed with which object strikes the ground is   30 m/s

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Answer 2

Given:

An object is dropped from the top of a tower. It

travels a distance 5 m in the first second and 25 m in the last second.

To find:

Velocity of striking the ground?

Calculation:

This question becomes very easy if you know the equation of DISPLACEMENT FOR nth SECOND.

• Let's say that the object travelled for t seconds.

• So it travelled 25 metres in the $t^{th}$ second.

Now, applying the required Equation:

$\sf \: s_{ {t}^{th} } = u + \dfrac{a}{2} (2t - 1)$

• Since object is dropped, value of 'u' is zero.

$\sf \implies\: s_{ {t}^{th} } = 0 + \dfrac{g}{2} (2t - 1)$

$\sf \implies\: 25= \dfrac{10}{2} (2t - 1)$

$\sf \implies\: 2t - 1 = 5$

$\sf \implies\: 2t = 6$

$\sf \implies\: t = 3 \: sec$

Now, total time for which object travelled is 6 secs.

• Let final Velocity of striking be v :

$\sf \: v = u + at$

$\sf \implies\: v = 0+ gt$

$\sf \implies\: v = gt$

$\sf \implies\: v = 10 \times 3$

$\sf \implies\: v = 30 \: m/s$

So, Velocity if striking is 30 m/s.