Question

A bullet is fired vertically upwards with an initial velocity of 50ms-1 .it covers a distance h1 during the first second and a distance h2 during the last 3 seconds of it's upwards motion .if g=10ms-2,h1 and h2 will be related as?

Answer 1

Given: A bullet is fired vertically upwards with an initial velocity of 50ms-1 .it covers a distance h1 during the first second and a distance h2 during the last 3 seconds of it's upwards motion.

To find: How is h1 and h2 related ?

Solution:

• Now we have 2nd equation of motion:

                s = ut + 1/2(at²)

                s = 50 + 1/2(-10)(1)^2

                s = h1 = 45 m

• Now time of ascent will be:

                t(a) = u/g = 50/10 = 5 second.

• so at t = 2s last 3 sec of upward motion starts and height will be:

                h = 50(2) - 10x4/2

                h = 100 - 20

                h = 80 m

                H(max) = u²/2g

                H(max) = 50²/2(10)

                H(max) = 2500/20

                H(max) = 125 m

• Now the distance traveled in last 3 sec h(2) will be:

                h2 = H - h

                h2 = 125 - 80

                h2 = 45 m

                Here h1 = h2

Answer:

         So the relation between h1 and h2 is h1 = h2.

Answer 2

Given:

A bullet is fired vertically upwards with an initial velocity of 50m/s. It covers a distance h1 during the first second and a distance h2 during the last 3 seconds of it's upwards motion.

To find:

Relation between h1 and h2.

Calculation:

Let total time for reaching maximum height be t.

$\therefore \: v = u + at$

$= > \: 0 = 50 + ( - 10)t$

$= > \: t = 5 \: sec$

So , body will reach maximum height in 5 seconds.

Displacement travelled in 1st second of ascent be h1.

$\therefore \: h1 = ut + \dfrac{1}{2} a {t}^{2}$

$= > \: h1 = (50 \times 1) - (\dfrac{1}{2} \times 10 \times {1}^{2} )$

$= > \: h1 = 50 - 5$

$= > \: h1 = 45 \: m$

Displacement travelled in the last 3 seconds of ascent be h2.

$\rm{h2 = displac. \: in \: 5 \: sec - displac.\: in \: 2 \: sec}$

$= > h2 = (50 \times 5 - \frac{1}{2} \times 10 \times {5}^{2} ) - (50 \times 2 - \frac{1}{2} \times 10 \times {2}^{2} )$

$= > h2 = 125- 80$

$= > h2 =45 \: m$

So , final answer is :

$\boxed{ \rm{ \red{h1 = h2 =45 \: m}}}$