Question

A Pebble thrown vertically upward with an initial velocity 500 m/s comes to a stop in 25seconds. Find the retardation. *​

Answer 1

Answer:

u = 500 m/s

v = 0 m/s

t = 25 s

Then,

a = v - u / t

a = 0 - 500 / 25

a = -500 / 25

a = -20 m s−2

Ans : Retardation = 20 m s−2

Explanation:

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Answer 2

Given that, a pebble thrown vertically upward with an initial velocity 500 m/s. Here, the initial velocity i.e. u is 500 m/s.

The pebble stops in 25 seconds. Here, time 't' is 25 seconds. Whereas the final velocity 'v' of the pebble is 0 m/s.

We have to find the retardation of the pebble.

Using the First Equation Of Motion,

v = u + at

From above data, we have v = 0 m/s, u = 500 m/s and t = 25 sec

If we substitute these values in the above we can find the value of acceleration.

→ 0 = 500 + a(25)

→ 0 = 500 + 25a

→ -500 = 25a

Divide by 25 on both sides

→ -500/25 = 25a/25

→ -20 = a

→ a = - 20

On solving we get acceleration = -20 m/s². (Negative sign shows retardation)

Therefore, the retardation is -20 m/s².