Question

An object from the rest covers the last 75% of it’s fall (With respect to height) in1sec. From what height was it dropped?

Answer 1

Answer:

19.62 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$\frac{s}{4}=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{4}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{4}=\frac{1}{2}at^2$

$\frac{3}{4}s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{3}{4}s=u1+\frac{1}{2}\times a\times 1^2\\\Rightarrow \frac{s}{4}=\frac{u}{3}+\frac{1}{6}a$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=\frac{u}{3}+\frac{1}{6}a\\\Rightarrow \frac{1}{2}at^2=\frac{at}{3}+\frac{1}{6}a\\\Rightarrow 3t^2-2t-1=0$

$t=\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:3\left(-1\right)}}{2\cdot \:3}, t=\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:3\left(-1\right)}}{2\cdot \:3}\\\Rightarrow t=1, -\frac{1}{3}$

Time taken to first 25% of the fall is 1 second

Total time taken = 2 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 9.81\times 2^2\\\Rightarrow s=19.62\ m$

The height from which the object was dropped was 19.62 m

Answer 2

By using the motion equations, we will see that the object is dropped from a height of 8.71 meters.

Finding the motion equations.

We should start with the acceleration, which will be equal to the gravitational acceleration, then we have:

a(t) = -9.8 m/s^2

To get the velocity equation we integrate over time, because the object is dropped, it does not have initial velocity, thus there is no constant of integration.

v(t) = (-9.8 m/s^2)*t

To get the position we integrate again, here the constant of integration is the initial height, then we have:

p(t) = (-4.9 m/s^2)*t^2 + H

The time in which the object falls to the ground is given by:

p(t) = 0 =  (-4.9 m/s^2)*t^2 + H

(4.9 m/s^2)*t^2 = H

t = √(H/4.9 m/s^2)

And we know that 75% of that is equal to 1s, then:

0.75*√(H/4.9 m/s^2) = 1s

Now we can solve this for H.

H = (4.9m/s^2)(1s/0.75)^2 = 8.71m

So the object is dropped from 8.71 meters.

If you want to learn more about motion equations, you can read:

https://brainly.com/question/11049671