Question

A stone is thrown vertically upward with an initial speed of 5 m/s. What is the velocity of the stone 3 seconds later?

Answer 1

-25m/s

Step-by-step explanation:

For motion under gravity:

I. For a body falling from a particular height the equations are:

a. V = u + gt

b. v2 = u2 + 2gh

c. h = ut + 1/2gt2

II. For a body thrown vertically upwards the equations become:

a. V = u – gt

b. v2 = u2 – 2gh

c. h = ut – 1/2gt2

Where v = final velocity in m/s, u= initial velocity in m/s, h= height in m, t= time in secs, g= acceleration due to gravity = 10m/s2.

Using,

v = u - gt

Where u = 5m/s, g = 10m/s2, t = 3s

v = 5 - 10 x 3

v = 5 - 30

v = -25m/s

The negative value of v means it's downward, that is, in the opposite direction of the motion.

Note that the approximate value of 'g' was taken as 10m/s2 since the value wasn't given. You should use the value of 'g' given you to solve.

Answer 2

`v_f=-24.4(m)/(s)`

Step-by-step explanation:

The acceleration due to gravity is in the downward direction. This means that it will oppose the upward motion of the stone. This will cause the stone to stop and begin to fall. Therefore, the sign of the final velocity is negative, since it is in the downward direction. The final velocity is given by:

`v_f=v_0+gt\\v_f=5(m)/(s)+(-9.8(m)/(s^2))(3s)\\v_f=-24.4(m)/(s)`