Question

an object thrown vertically upward reaches a height of 500m what was its initial velocity? how long will be the object take to come back to the earth? assume g=10 m/s²


Please solve it step-by-step

Answer 1

Hi friend
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Your answer
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Height attained or distance covered in upward direction (s) = 500 m

Acceleration due to gravity (g) = - 10m/s² [The body is going in a direction opposite to gravity ,so acceleration is taken negative.]

Final velocity (v) = 0 m/s. [Because the object will stop after reaching the height of 500 m .]

Initial velocity with which it was thrown up (u) = ?

Using formula : - v² - u² = 2as

(0)² - u² = 2 × (- 10) × 500

=> - u² = - 10000

=> u² = 10000

=> u = √10000

=> u = 100 m/s

Now,
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Time taken to come back to earth (t) = ?

In this case,

u = 0 m/s

a = 10 m/s²

using formula : - s = ut + ½ × a × t²

=> 500 = (0 × t) + ½ × 10 × t²

=> 500 = 0 + 5t²

=> t² = 500/5

=> t² = 100

=> t = √100

=> t = 10 m/s

HOPE IT HELPS

Answer 2

let K = kinetic energy
P = potential energy

initially
K = 1/2mv²

and
finally
K = 0

by energy conservation
change in K = change in P

so
1/2mv² = mgh
1/2v² = gh

v = √2gh


so
v = √(2×10×500)
v = √10000
v = 100 m/s

initial velocity = 100m/s


by the formula
v = u + at
100 = 0 + gt
t = 100/g
t = 10 sec

so
time to come back to the earth
t = 10 second

hope it helps you
@di