Question

. A ball is thrown straight up with a velocity of 14m/s. What will be the velocity 3 seconds after being released?

Answer 1

Explanation:

so given initial speed is,u = 14m/s

and acceleration due to gravity is,g = 9.8m/s^2

and given time is,t = 3 seconds

we have to find final velocity 3 seconds after being released, v = ?

using kinematics

v = u - g*t

we are using negative sign for 'g' because ball going against the gravitational force

v = 14m/s - (9.8m/s^2)*3s

v = 14m/s - 29.4m/s

v = - 15.4 m/s

so final velocity of the ball after 3 seconds is 15.4m/s

negative sign showing that ball going to upward

Answer 2

Explanation:

using Newton's first equation of motion

v=u+at

now let v=0(I have kept V's value to find the maximum time from where the ball starts coming down)

0=14-10t

t=1.4 seconds

so at the end of 1.4 seconds ball's final velocity is zero

total time is 3

maximum time 1.4 seconds

time from where ball starts coming down

3-1.4=1.6

again using Newton's first equation of motion with time's value as 1.6

v=u+at

v=0+10*1.6

v=16m/s