Question

A stone is dropped from the top of a tower and it is found to travel 7/16 of height of tower during the last second of its fall . The height of tower is (g=10 m/s^2)

Answer 1

Answer:

Explanation:

Using 4th equation of motion,

Distance in last second = a(t-0.5)

take total distace as ½gt² = 5t²

7/16h = 35/16t²

So, 35/16t² = 10t - 5

=> 35t² = 160t - 80

=> 35t² - 160t +80 =0

=> 7t² - 32t +16 = 0

This is a quadratic equation in t.

By solving, we get t = 4seconds.

So height of tower is 5t² = 5 x 16 = 80m

Answer 2

The height of the tower is equal to 80 metres.

Given:

The stone travels 7/16 of the height of the tower during the last second of its fall.

Acceleration due to gravity (g) = 10 ms⁻²

To Find:

The height of the tower.

Solution:

Let the height of the tower be 'H'.

The initial velocity of the stone (u) = 0

Let the total time interval of the free fall be 'T'.

→ The stone is travelling a distance of 'H' in 'T' seconds.

∵ The stone travels a distance of H(7/16) during the last second.

∴ Distance travelled by the stone in (T - 1) seconds= H - (7/16)H = H(9/16)

→ By using the 2nd Equation of Motion:

→ The stone is travelling a distance 'H' in 'T' seconds:

   $\therefore S=ut+\frac{1}{2}at^{2} \\\\\therefore H =0(T)+\frac{1}{2}gT^{2} \\\\\therefore H=\frac{1}{2}gT^{2} -Eq.(i)$

→ The stone is travelling a distance H(9/16) in (T - 1) seconds:

   $\therefore S=ut+\frac{1}{2}at^{2} \\\\\therefore \frac{9}{16}( H) =0(T)+\frac{1}{2}g(T-1)^{2} \\\\\therefore \frac{9}{16} H=\frac{1}{2}g(T-1)^{2} \\\\\therefore 9H=8g(T-1)^{2} -Eq.(ii)$

→ Dividing equation (ii) by equation (i):

   $\therefore9=\frac{8g(T-1)^{2} }{(1/2)gT^{2} } \\\\\therefore 9gT^{2} =16g(T-1)^{2} \\\\\therefore 9T^{2} =16T^{2} +16-32T\\\\\therefore 7T^{2} -32T+16=0\\\\\therefore 7T^{2} -28T-4T+16=0\\\\\therefore 7T(T-4)-4(T-4)=0\\\\\therefore T=4\:seconds\:or\:4/7\:seconds$

→ But as we know the total time 'T' must be greater than 1 second. Therefore, 'T' must be equal to 4 seconds.

              $\therefore \fbox{T=4\:seconds}$

→ Putting the value of 'T' in equation (i):

    $\therefore H=\frac{1}{2}gT^{2} \\\\\therefore H=\frac{1}{2}\times10\times(4)^{2} \\\\\therefore H = 5\times16\\\\\therefore H=80\:metre$

Therefore the height of the tower is equal to 80 metres.

#SPJ2