Question

An object thrown vertically upwards reaches a height of $500m$. What was its initial velocity? How long will the object take to come back to the earth? Assume $g=10m/s^2$

Answer 1

Step 1: Given that:

The height(h) reached by the object thrown upward = 500m

Acceleration due to gravity(g) = $10m{s}^{-2}$

Step 2: Calculation of initial velocity given to the object:

On the vertical upward motion of a body, the acceleration due to gravity is taken as negative.

At maximum height, the final velocity(v) of the object is zero.

Now, using third equation of motion under gravity for the upward motion of a body;

$v^2=u^2-2gh$

Here, v= final velocity, u= initial velocity, g = acceleration due to gravity and height attained by the object

Putting the given values we get;

$0=u^2-2\times 10m{s}^{-2}\times 500m$

$u^2=10000$

$u=100m{s}^{-1}$

Step 3: Calculation of the time taken by the object to come to the surface of the earth:

Using, the first equation of motion under gravity for upward motion,

$v=u-gt$

$0=100m{s}^{-1}-10m{s}^{-2}\times t$

$10t=100$

$t=10sec$

Since this is the time to reach at maximum height.

The object will take same time to reach to the surface of the earth from that height.

Therefore net time= $10sec+10sec=20sec$

Thus,

The initial velocity of the object = 100m/sec

Time taken to reach to the surface of the earth = 20sec