A ball is thrown straight up whith an inital speed of 30m/s how high does it go and how long is it in the air?

Question

(neglecting air resistance)

taking in to account linear motion

Answer 1

using;

v^2(final) = v^2(initial) +2as

0 = 30^2 + 2 (-9.81)s

solve for s....s=45.87m

Answer 2

Height reached= (30^2)/2g=900/20=45 metres. Time taken to reach this height=30/g=3 seconds. It will also take 3 seconds to fall to the ground again, so it will be in the air for 6 seconds.

Answer 3

Using:
s = u.t + 0.5 * a t^2
(s=displacement, U= Initial Velocity, a=acceleration, t=time)
Since we want the time it was in the air, we put 0 as displacement (It goes up and falls down to the same place, so s = 0)
0 = 30t + 0.5 * (-9.81) * t^2

the answer gives;
t=0 or t=6.11

Both are correct since the ball is on ground initially and at 6.11s.
The answer you want is 6.11s.

Answer 4

height=45mts
time=3sec