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

for 1st part use v^2=2gh
v^2=2*10*500
v=100 m/s
for 2nd part
v=at
100=10t
t=10sec
since it is only upgoing time therefore for total journey it is 20 sec

Answer 2

Answer :-

Magnitude of force is 7.5 Newtons .

Explanation :-

We have :-

→ Initial velocity (u) = 0 m/s

→ Mass of the body (m) = 3 kg

→ Time for action of force (t) = 2 sec

→ Distance (after force ceases) = 5 m

→ Time (after force ceases) = 1 sec

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From the question, we can understand that a force acts on a body for 2 seconds. After the force ceases, the body moves with constant speed (as there is no acceleration) for 1 second.

So, speed of the body after action of force :-

⇒ Speed = Distance/Time

⇒ v = 5/1

⇒ v = 5 m/s

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Now, we will calculate the acceleration of the body for the time when force was acting .

v = u + at

⇒ 5 = 0 + a(2)

⇒ 5 = 2a

⇒ a = 5/2

⇒ a = 2.5 m/s²

Finally, we can calculate the magnitude of force by putting obtained values in Newton's 2nd law of motion .

F = ma

⇒ F = 3 × 2.5

⇒ F = 7.5 N