Asked by Carmieee

A ball is thrown up in the air with an initial velocity of 15m/s from a 17m high platform

Find
Maximum height reached by the ball
Time of flight if the ball is catched
Time of flight when it strike the ground
Velocity when it returns to the hand
Velocity when it strike the ground
Answered by bobpursley
I will be happy to critique your thinking on this.
Answered by Damon
a = - g = -9.81 m/s^2
v = 15 - 9.81 t
v = 0 at top so solve for t at top

H = max height = 17 + 15 t - 4.9 t^2

now it falls from height H
17 = H - 4.9 t'^2
solve for t' , the fall time from top to 17, v here = -9.81 t' add t to t' to get flight time so far

0 = H -4.9 t"^2
solve for t", the time to fall from top to ground
v at ground = -9.81 t"
time of flight = t + t"
Answered by Carmieee
My max height results into 28.47m
I used the equation
Vf^2=Vo^2+2gdy

In the time of flight when it reaches the hand will be 1.53s
Using the eqn
Gt=Vf-Vo

Time as it strikes the ground will be 3.94s
Using the eqn
dy=Vot+.5(-9.81m/s^2)t^2
Answered by Damon
suggest you check by doing it my way.
Answered by Carmieee
May I ask if t= time of flight going up and t"=time of flight going down
Answered by Carmieee
Can I really solve this without the magnitude of the hand from the ground which considered to be the reference point?
Answered by Carmieee
I think it lacks information when it comes to solving for time and final velocity as it reaches the hand
Answered by Damon
assume hand at 17 meters height.

t = rise time indeed

t' = time to fall from top to 17 meters

t" = time to fall from top to ground
Answered by Carmieee
I got same value for t' and t is it possible? And if I'm about to add it it will be t"
Answered by Carmieee
Another question is in finding the final velocity.
Using the gt=vf-vo

Will the values depend on the time fall as it reach the coin and is it the initial velocity in the top also 0m/s?
Answered by Damon
at the top the velocity is indeed zero.

so if you know the fall time, t"
then v at the ground = - g t"
the - sign is because v is down.

YES, t = t' by symmetry
it takes just as long to deaccelerate to the top as to accelerate down to the starting point.

NO, t" includes the time from 17 meters up to the ground
Answered by Carmieee
Okay I got it now. Thank you so much sir! :)
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