Science, asked by Anshukumar111, 1 year ago

A ball is dropped from a tower .in the last second of it motion it travels a distances of 15m . find the hight of tower .

Answers

Answered by AR17
302
Heya user !!

Here's the answer you are looking for

Let the height of the tower is h and time taken by it to reach the ground from height h is t.

Since, it is dropped, it's initial velocity (u) = 0

Using the second equation of motion,

s = ut + \frac{1}{2} g {t}^{2} \\ h = 0 + \frac{1}{2} (10) {t}^{2} \\ h = 5 {t}^{2} \\ t = \sqrt{ \frac{h}{5} }

The last second means at t = √(h/5)th second.

So, at t = √(h/5)th second, the ball travels 15m

 s_{nt} = u + \frac{a}{2} (2n - 1) \\ 15 = 0 + \frac{10}{2} (2 \: \sqrt{ \frac{h}{5} } - 1) \\ 15 = 5(2 \: \sqrt{ \frac{h}{5} } - 1) \\ \frac{15}{5} = (2 \: \sqrt{ \frac{h}{5} } - 1) \\ 3 = 2 \: \sqrt{ \frac{h}{5} } - 1 \\ 3 + 1 = 2 \: \sqrt{ \frac{h}{5} } \\ \frac{4}{2} = \sqrt{ \frac{h}{5} } \\ 2 = \sqrt{ \frac{h}{5} } \\ \\ squaring \: both \: sides \\ \\ 4 = \frac{h}{5} \\ h = 4 \times 5 = 20

Therefore, the height of the tower is 20m.

★★ HOPE THAT HELPS ☺️ ★★

Anshukumar111: Thanku
AR17: no problem!! :-)
Answered by syed2020ashaels
2

Answer:

The height of tower is 20 m

Explanation:

  • Now as we know that the equation of motion is
    s=u*t+\frac{1}{2} * a* t^2
    and for nth second we have the equation of motion as
    s_{n}=u*t_{n}+\frac{1}{2} * a* (2*t_{n} - 1)
  • So by assuming height of the building to be h and total time taken by the ball to reach ground be t and initial velocity is 0 hence we get the equation as
    h = 0*t+\frac{1}{2}*10*t^2;
    on solving this we get
    \frac{h}{5}= t^2\\t=\sqrt{\frac{h}{5} }
  • Now in last second that is \sqrt{\frac{h}{5}} th second it travelled 15 m so
    15 = 0 + \frac{1}{2} * 10 * (2*\sqrt{\frac{h}{5}} -1)
  • After solving this we get the value of h as 20m
    #SPJ2
Similar questions