A ball is thrown vertically upward with a
speed of 25.4 m/s from a height of 2.1 m. (Remember gravity is a=9.8m/s^2)
How long does the ball take to hit the ground after it reaches its highest point?
Answer in units of s.
3 answers
make a parobala, find the highest point find the time for the whole thing, then dive the time in half. hope that kind of helped !
height Y = 2.1 + 25.5 t - 4.9 t^2
Solve for the time Y = 0
It reaches its highest point when t = Vo/g = 25.4/9.8 = 2.59 s
Since they want the time it takes to hit the ground AFTER reaching the highest point, you will need to subtract 2.59s from the time Y=0
There is another way to do this without solvoing a quadratic equation>
1) use conservation of energy to calculate how much higher high it goes.
g*(deltaY) = Vo^2/2
2) Add the initial elevation, 2.1 m, to delta Y. That will give you Ymax
3) T (time to fall from max height)
= sqrt (2 Ymax/g)
Solve for the time Y = 0
It reaches its highest point when t = Vo/g = 25.4/9.8 = 2.59 s
Since they want the time it takes to hit the ground AFTER reaching the highest point, you will need to subtract 2.59s from the time Y=0
There is another way to do this without solvoing a quadratic equation>
1) use conservation of energy to calculate how much higher high it goes.
g*(deltaY) = Vo^2/2
2) Add the initial elevation, 2.1 m, to delta Y. That will give you Ymax
3) T (time to fall from max height)
= sqrt (2 Ymax/g)
1,\m/s
1 answer