A ball is thrown straight up with an initial speed 40m/s. How high does it go and how long is in the air?

Question

Please include formulas and remember (neglecting air resistance)

Answer 1

learned this first week of freshman HS physics

Acceleration due to gravity is 9.8 m/s^2

I assume your teacher is rounding to 10m/s^2

so if it is thrown straight up at 40m/s, every second it will lose 10m/s of speed
so at the end of the first second it will be going 30m/s
2nd = 20m/s
3rd = 10m/s
4th = 0m/s
5 = -10m/s
6 = -20m/s
7 = -30m/s
8 = -40m/s

Answer 2

Answer is 80 m, and 8 s.

You need to know there are five variables s u v a t

And four equations taking any four out of the five variables.

v = u + at
s = (u + v)t /2
v<2> = u<2> + 2as
s = ut + at<2>/2

Answer 3

v^2=2as, so s=(v^2)/2a=1600/20=80 metres. This is the height it reaches. s=1/2 at^2 so t= SQR(2s/a)=SQR(160/10)=4; so it takes 4 seconds to climb to maximum height. It will take another 4 to fall to the ground so it is in the air 8 seconds.

Answer 4

v^2-u^2=2as
here v=o
Therefore s=u^2a (-ve signs cancel out)
or s=80 m taking g+10m/s^2
v=u+at
ie t=u/a=4s

Answer 5

g is 9.8 meters per second each second;
velocity = space : time;
try to get your Physics text and go on. This space is not for avoid study in school...................................................... i think.