Question

A ball released from the top of a tower travels 11/36 of the height of the tower in the last second of its journey.The height of the tower is(take g=10m/s^2)

Answer 1

Solution:

According to third kinematic equation of motion,

d = ut+$\frac{1}{2} at^{2}$

Where d = distance

u = initial velocity = 0

a = acceleration = acceleration due to gravity = 10 m/$s^{2}$

Thus, Total distance is given by,

d = $\frac{1}{2} gt^{2}$...(1)

Now Lets assume distance traveled in t-1 second is $d_{1}$

$d_{1}$ = $\frac{1}{2} g(t-1)^{2}$...(2)

Subtracting equation (2) from equation (1), we get

d - $d_{1}$ = $\frac{1}{2} gt^{2}$ - $\frac{1}{2} g(t-1)^{2}$...(3)

distance traveled in last second is $\frac{11}{36}d$

Thus, equation (3) becomes

$\frac{11}{36}d$ = $\frac{1}{2} gt^{2} - \frac{1}{2} g(t-1)^{2}$

$\frac{11}{36}d$ =$\frac{1}{2} gt^{2} - \frac{1}{2}g(t^{2} - 2t + 1)$

From equation (1), we get

$\frac{11}{36}\frac{1}{2} gt^{2}$ =$\frac{1}{2} g(2t-1)$

$\frac{11}{36} t^{2}$ =$(2t-1)$

$11t^{2} -72t + 36 = 0$

on solving above equation we get,

t= $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

t= $\frac{72\pm 60}{22}$

t= 6 second

Now height is

d = $\frac{1}{2} gt^{2}$

d = 176.40 m

Answer 2

Answer:

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Explanation: