Vi = 30
v = Vi - 9.81 t
at top v = 0
so t at top = 30 / 9.81
50 + 30 t - 4.9 t^2
t plus time needed to fall from part B height
0 = 30 + Vi(total time) - 4.9 total time^2
solve for total time
v = Vi - gt the whole time
h = 30 + Vi t - 4.9 t^2
A ball is thrown from the top of a building with an initial velocity of 30m/s straight upward, at an initial height of 50m above the ground. The ball just misses the edge of the roof on is way down and hits the ground. Determine: a) time needed for ball to reach max height. B)the max height. C) time needed for ball to hit ground. D) velocity and diplacement ( magnitude and direction) of the ball at t=sec( measured from beginning. Please explain step by step
6 answers
What did you use to find the time for the egg to reach it's maximum height? The answer that you put is kinda confusing.
v = Vi - g h
v is 0 at top
so 0 = Vi - 9.81 t
or t = Vi/9.81 This should be embedded in your soul
v is 0 at top
so 0 = Vi - 9.81 t
or t = Vi/9.81 This should be embedded in your soul
You need it for every problem like this
remember, vertical velocity is zero at the top
and horizontal velocity is constant the whole time
remember, vertical velocity is zero at the top
and horizontal velocity is constant the whole time
I don't understand the maximum height part? what is the formula?
Hey, you throw something straight up at velocity +Vi
The force on it is down, -m g
F = m a
-m g = m a
so
a = -g
so
v(t) = -g t + a constant
since v(0) = Vi the constant is Vi
v(t) = Vi - g t
NOW when it gets to the top, the velocity v( t at top) = 0
so at top
0 = Vi - g t
SO
t = Vi/g AT THE TOP
then for height
h = Hi + Vi t - 4.9 t^2
The force on it is down, -m g
F = m a
-m g = m a
so
a = -g
so
v(t) = -g t + a constant
since v(0) = Vi the constant is Vi
v(t) = Vi - g t
NOW when it gets to the top, the velocity v( t at top) = 0
so at top
0 = Vi - g t
SO
t = Vi/g AT THE TOP
then for height
h = Hi + Vi t - 4.9 t^2