A ball is thrown vertically upwards with an initial velocity of 27.0 m/s. Neglecting air resistance, how long is the ball in the air? What is the greatest height reached by the ball? And calculate the first time when the ball has half of its initial velocity.

Thank you!

v =u-gt wher as v= final vilocity= 0 (at top point ), u =initial vilocity =27.0m/s,

g= gravitational force =9.8m/s
time in air is
t=2(u/g)

v2=u2-gt


at half of vilocity that is v=u/2=27/2
v =u-gt
solve this eqution get your answer

The formula is

The initial velocity being 27m/s, the final velocity will be -27m/s when the ball comes down at the same height as it has been thrown. The acceleration is negative since it is in the opposite direction to your initial velocity. So, set u to 27, a to -9.81 and v to -27. Solve for t to get the time the ball was in the air. (ans: 5.5 s)

Use to find the max height.

The initial velocity is 27 m/s, the final velocity is 0 m/s (since it is at the maximum height), and the acceleration is -9.81 m/s^2 (since it is again in the opposite direction to your initial velocity). Solve for s, the displacement. (ans: 37.2m)