a ball thrown upward with a velocity of 19.6m/s will come momentarily in rest in?
Ans 2sec
time to peak (zero velocity) ... v / g = 19.6 m/s / 9.8 m/s^2
thnx
To determine the time at which the ball thrown upwards with a velocity of 19.6 m/s comes momentarily to rest, we can use the concept of motion under gravity.
1. Identify the initial velocity (u) of the ball as 19.6 m/s. Since the ball is thrown upwards, the initial velocity will be positive.
2. The acceleration due to gravity (g) acts in the opposite direction of the motion, so it will have a negative value of approximately -9.8 m/s².
3. The final velocity (v) of the ball will be zero when it momentarily comes to rest.
4. Use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since we want to find the time (t), we rearrange the equation to solve for t:
t = (v - u) / a
Substituting the known values into the equation:
t = (0 - 19.6) / (-9.8)
t = 19.6 / 9.8
t = 2 seconds
Therefore, the ball will come momentarily to rest after 2 seconds.