Question

In a rescue operation a food packet is dropped from
a height of 500 m. Find time taken by it to reach
the ground. Also calculate its velocity with which it
hits the ground. Take g = 10 m/s2​

Answer 1

Answer:

With calculus, huh. We know the vertical acceleration is a constant −g.

d2ydt2=−g

Integrating, we’ll call our constant v0.

dydt=v0−gt

That’s our vertical velocity as a function of time. In our particular case we can say the initial velocity v0=0 , i.e. we started from rest at time t=0.

dydt=−gt

Let’s integrate again, integration constant y0 .

y=y0−12gt2

We have y0=500 , and we want t when y=0.

0=500−12gt2

gt2=1000

t=1000g−−−−−√

We often use the approximation g=10m/s2 which here gives t=10

What speed is it going at t=10 ?

dydt=−gt=−(10)(10)=−100