Question

1.a body travels a distance of 20m in 7th second and 24m in the 9th second .how much distance shall it travel in the 15th second?
2.a body falls freely from the top of the tower .it covers 36 per cent of the total height in the last second before striking the ground level .find the height of the tower ?

Answer 1

Answer:

Step by step:

Thanks

Answer 2

Answer:

1. Height of the tower is 36m.

2. The height of the tower = 125m

Explanation:

Assume:

Body is traveling with uniform acceleration= a

Initial velocity= u,

The distance traveled at 7th = S₇

Distance travelled at 9th second = S₉

Distance traveled by the body at the nth second=$S_{n} = u+\frac{1}{2} a(2n-1)$

Using this relation for 7th and 9th second, we get

n=7

$S_{7}=U+\frac{1}{2} a(2*7-1)=20\\\\ =U+\frac{1}{2} 13a=20$....................................(1)

n=9

$S_{9}=U+\frac{1}{2} a(2*9-1)=24\\\\ =U+\frac{1}{2} 17a=24$.......................................(2)

Hence by subtracting equations 1&2 we get,

4=2a..............................................(3)

Therefore the acceleration a=2ms⁻².

Substituting this in the equation 1 we get,

        20=u+(12×2×(13))

The initial velocity u=7m/s.

Hence the distance traveled in the 15th

second is given by $S_{15}=7+\frac{1}{2} *2(2*15-1)=36m$

Thus the body will travel 36m in the 15th second.

question:2

Let height of tower = h

body takes t time to reach to ground when it fall freely.

Then initial velocity=u

S=ut+1/2gt²

here s=h

U=0 (INITIAL AT REST)

Therefore $h=\frac{1}{2} g t^{2} \ldots \text {. (i) }$

In last second i.e., $t^{th}$ second body travels =36%h=0.36 h

It means in rest of the time i.e., in (t-1) second it travels =h-0.36

h=0.64 h

Now applying equation of motion for t-1 second

$0.64 h=\frac{1}{2} g(t-1)^{2}$

From Eqs. (i) and (ii) we get, t=5 $\mathrm{~s} and \mathrm{h}=125 \mathrm{~m}$

The height of the tower =125m