Question

I throw a ball in the air at 10m/s. I want to measure xmax which is the highest point that ball will reach. I assume gravity is 10m/s^2, and I assume there is no air resistance. I get xmax= x0 +v0*t1-((g*t1^2)/2) [x0=starting height(lets say it's 0m), v0= speed at which I threw the ball (10m/s), g=gravity(10m/s^2), t1= time it takes the ball to reach xmax). What is the logic behind ((g*t1^2)/2)? I understand gravity reduces the speed of the ball but why is it multiplied by (t1^2)/2

Answer 1

i guess there is a huge derivation about it .... so i guess it must have been given a good explantion in your book

Answer 2

\[Xmax=X_{0}+V_{0}*t _{1}-(\frac{ g*(t _{1})^2 }{ 2 } )\]

Answer 3

\[X _{0}= 0m; V _{0}=10\frac{ m }{ s }; g=10\frac{ m }{ s^{2} }\]

Answer 4

t 1=time it takes the ball to reach Xmax