Question

An object thrown vertically upwards
reaches a height of 500 m. What was
its initial velocity? How long will the
object take to come back to the
earth? Assume g = 10 m/s2​

Answer 1

Explanation:

14m/s,1.43s

Solution :

Here, h=10,v=0, at highest point, u=?,t=?,g=−9.8m/s2

From v2−u2=2gh,0−u2=2(−9.8)10=196 or u=196−−−√=14m/s

v=u+gt,0=14−9.8tt=149.8=1.43s

answer

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)2gs=v

2

−u

2

→(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}2×(−10)×500=0−u

2

Thus,

10000=u^{2}10000=u

2

Hence, the initial velocity is

u=100\ \mathrm{m} / \mathrm{s}u=100 m/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)v=u+at→(2)

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

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

0=100−10t

10t=100

t=10s

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.

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