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
I will be happy to critique your thinking on this.
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"
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
suggest you check by doing it my way.
May I ask if t= time of flight going up and t"=time of flight going down
Can I really solve this without the magnitude of the hand from the ground which considered to be the reference point?
I think it lacks information when it comes to solving for time and final velocity as it reaches the hand
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
I got same value for t' and t is it possible? And if I'm about to add it it will be t"
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?