Question

A ball with mass 0.11 kg is thrown upward with initial velocity 10 m/s from the roof of a building 20 m high. Neglect air resistance. (A computer algebra system is recommended. Use g = 9.8 m/s2 for the acceleration due to gravity. Round your answers to one decimal place.)
(a) Find the maximum height above the ground that the ball reaches.

(b) Assuming that the ball misses the building on the way down, find the time that it hits the ground.

(c) Plot the graph of velocity versus time.

(d)Plot the graph of position versus time.

Answer 1

Answer:

a) $H=25.1020\ m$

b)  $t=3.2838\ s$

Explanation:

Given:

• mass of the ball thrown up, $m=0.11\ kg$

• initial velocity of the ball thrown up, $u=10\ m.s^{-1}$

• height above the ground from where the ball is thrown up, $h=20\ m$

a)

Maximum height attained by the ball above the roof level can be given by the equation of motion.

As,

$v^2=u^2-2g.h'$

where:

$v=$ final velocity at the top height of the upward motion $=0\ m.s^{-1}$

$g=$ acceleration due to gravity

$h'=$ height of the ball above the roof

Now,

$0^2=10^2-2\times 9.8\times h'$

$h'=5.10\ m$

Therefore total height above the ground:

$H=h+h'$

$H=20+5.1020$

$H=25.1020\ m$

b)

Now we find the time taken in raching the height $h'$:

$v=u-gt'$

$v=$ final velocity at the top of the motion $=0\ m.s^{-1}$

So,

$0=10-9.8\times t'$

$t'=1.0204\ s$

Now the time taken in coming down to the ground from the top height:

$H=u'.t_d+\frac{1}{2} g.t_d^2$

where:

$u'=$ is the initial velocity of the ball in course of coming down to ground from the top $=0\ m.s^{-1}$

Here the direction acceleration due to gravity is same as that of motion so we are taking them positively.

$25.1020=0+0.5\times 9.8\times t_d^2$

$t_d=2.2634\ s$

Therefore the total time taken in by the ball to hit the ground after it begins its motion:

$t=t'+t_d$

$t=1.0204+2.2634$

$t=3.2838\ s$