Question

an object throw 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=10 m/s square

Answer 1

Hello Mate!

v^2 - u^2 = 2as
0 - u^2 =2 × 10 × 500
-u^2 = 10000
u= -100 m/s

v - u = at
( v - u ) / a = t
( 0 - ( -100 ) ) / 10 = t
10 seconds = time

Hope it helps☺!

Answer 2

Answer:  

The initial velocity is 100 m/s and the total time taken to reach the ground is 20 s.

Solution:

From Newton’s third equation of motion:

$2 g s=v^{2}-u^{2} \rightarrow(1)$

As the final velocity of the object when it reaches the ground will be zero, v =0 and acceleration is acceleration due to gravity during free fall. Then, the initial velocity to cover a distance of 500 m from ground is  

$2 \times(-10) \times 500=0-u^{2}$

Thus,

$10000=u^{2}$

Hence, the initial velocity is  

$u=100\ \mathrm{m} / \mathrm{s}$

Using the initial and final velocity, we can first determine the time taken by the object to reach the height of 500 m using Newton’s first equation of motion.

Now consider the Newton’s first equation of motion:

$v=u+a t \rightarrow(2)$

Now substitute the value of u in equation (2), we get

$\begin{array}{c}{0=100-10 t} \\ {10 t=100} \\ {t=10 s}\end{array}$

Thus, 10 s is required by the object to reach the from ground to 500 m height then same 10 s will be required by the object to again reach the ground. Thus the total time taken to reach the ground will be 20 s.