A ball is thrown vertically upwards with a velocity of 20 ms^(-1) from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise ? and (b) how long will it be before the ball hits the ground ? Take g = 10 ms^(-2).

A ball is thrown vertically upwards with a velocity of 20 ms^(-1) from

1.	A ball is thrown vertically upwards with a velocity of 20 ms^(-1) from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise ? and (b) how long will it be before the ball hits the ground ? Take g = 10 ms^(-2).
Answer» SOLUTION :(a) Let us take the y-axis in the vertically upward direction with zero at the ground, as shown in Fig. Now`upsilon_(0)=+20 ms^(-1)`, `a=-g=-10 ms^(-2)`, `upsilon = 0 ms^(-1)` If the ball RISES to height y from the POINT of launch, then using the equation `upsilon^(2)=upsilon_(0)^(2)+2(y-y_(0))` we get `0=(20)^(2)+2(-10)(y-y_(0))` Solving, we get,`(y-y_(0))=20m`. (b)We can solve this part of the problem in two ways.