Question

A)Calculate the initial momentum of the ball? b)Calculate the time taken for the ball to reach a maximum height? c) The boy catches the ball. Construct a sketch of a velocity-time graph to illustrate the motion of the ball.

A boy throws a ball of mass 0.01 kg vertically upwards with a speed of 12ms^-1
g=10 ms^-1
d) State how you would find the maximum height reached by the ball from the velocity-time graph
e) Determine the maximum vertical height reached by the ball

Answer 1

Please see below

Explanation:

So,the equation of motion relating velocity and time of the ball is `v=12-10t` (using, `v=u-g t`,here initial velocity =`12ms^-1` and `g=10`) Considering upward direction to be positive

So,at the highest point of its motion,the ball comes to momentary rest,so we need to calculate using this equation the time after which `v` becomes `0`

So,we can write,

`0=12-10t`

or, `t=1.2s`

So,in this time the ball reaches its maximum vertical height.

Now,see the graph,we know area under the curve of velocity - time graph gives displacement,so we need to calculate displacement from `t=0` to `t=1.2s` as during this time it reached the maximum height.

So,the maximum height is `1/2 *12*1.2=7.2m`

And also see from the equation velocity again becomes `12ms^-1` after `2.4s` this is the time after which the ball returns to its point of projection.(put `v=-12` so we get, `-12=12-10t` or, `t=2.4s`)

And,initial momentum of the ball was `m*v=0.01*12=0.12Kgms^-1` directed upwards.