Question

To find the height of an overhead power line you throw a ball straight upward. The ball passes the line on the way up after 0.75 s, and passes it again on the way down after a total of 1.3 s.
a.) What is the height of the power line?
b.) What is the initial speed of the ball?

Answer 1

u throw the ball straight upward . Let the height of overhead power line be 'h' & inital speed of the ball be 'u'.

As during the motion of the ball the only force acting on the ball was gravity thus we can assume that the ball was moving upwards with constant acceleration -g.

I assumed the upward direction of motion to be positive & downward direction to be negative & that is why i gave negative sign to 'g' becoz it is acting downwards.

Answer 2

after t=0.75 sec net displacement of ball is h

so
h=0.75u-\(\frac{1}{2}g(0.75)^2\)......(1)

after t=1.3 sec also the net displacement is h

so

h=1.3u-\(\frac{1}{2}g(1.3)^2\).............(2)