A piece of stone is thrown vertically upwards. It reaches the maximum height in 3 seconds. If the acceleration of the stone be $9.8 m / s^{2}$ directed towards the ground, calculate the initial velocity of the stone with which it is thrown upwards.

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Solution

Given :

Final velocity v = 0 m/s

Gravity acceleration g = 9.8 m/s^2

time t = 3 sec

Then the calculate of initial velocity u from 1st equation of motion

v = u + gt

0 = u + 9.8 × 3

u = – 29.4 m/s

Negative sign shows direction of initial velocity is opposite the gravity acceleration.

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A piece of stone is thrown vertically upwards. It reaches the maximum height in 3 seconds. If the acceleration of the stone be 9.8m/s2 directed towards the ground, calculate the initial velocity of the stone with which it is thrown upwards.

Given : Final velocity v = 0 m/s Gravity acceleration g = 9.8 m/s^2 time t = 3 sec Then the calculate of initial velocity u from 1st equation of motion v = u + gt 0 = u + 9.8 × 3 u = – 29.4 m/s Negative sign shows direction of initial velocity is opposite the gravity acceleration.

A piece of stone is thrown vertically upwards. It reaches the maximum height in 3 seconds. If the acceleration of the stone be $9.8 m / s^{2}$ directed towards the ground, calculate the initial velocity of the stone with which it is thrown upwards.

Given :

Final velocity v = 0 m/s

Gravity acceleration g = 9.8 m/s^2

time t = 3 sec

Then the calculate of initial velocity u from 1st equation of motion

v = u + gt

0 = u + 9.8 × 3

u = – 29.4 m/s

Negative sign shows direction of initial velocity is opposite the gravity acceleration.