Solving a Vertical Ball Throwing Problem: Velocity and Time Calculation

[inbetween]

Simple Question (Stuck!)

Stuck on my christmas revision questions. Basically here it is below.

Homework Statement

A ball is thrown vertically with a speed of 17ms-1. A man standing upon a tower 5m above the thrower attempts to catch the ball but fails. He does however catch the ball successfully on its way down.

What is A) The balls velocity when he catches it, B) The time it takes between throwing the ball and it being caught.

Homework Equations

v^2 = u^2 + 2as

s=distance(displacement)
v=final velocity
u=initial velocity
a=acceleration
t=time

The Attempt at a Solution

I attempted to work out the maximum height reached by the ball as it passes the tower so i need to know how high it goes to work out the final velocity when it returns to be caught. I worked out the height reached as 14.73m and then took away the difference between the height of the tower (5m) and the maximum height giving the distance the ball falls from the reached height to the point where it is caught 5m above the thrower. Using this i have an initial velocity of zero and an acceleration as a result of gravity (9.8ms). I'm a bit stuck though where to go from here and not fully confident i am doing the solution correct in the first place.

 

[Doc Al]

What you've done so far will enable you to find the speed after falling from it's maximum height, using the equation you have provided.

As far as figuring out the time involved, look for another kinematic formula. How would you express the height as a function of time for this projectile?

 

[faiz4000]

i am telling you how to solve the 2nd part, while going up, use v=u+at where u would be 17ms-1 and v=0 and a=-9.8ms-2. You will get the time as 1.73s. Then while the ball comes down, put u=0 and v=13.51ms-2(the velocity you got in the 1st part of the question) and a=9.8ms-2 in the above formula to get the time it takes for the ball to come down. Add the two times and you get your answer