Question

a boy p stands on top of a tower another boy q throws a ball vertically up from the ground so that it just reaches p the time taken is t0 the speed with which p should throw the ball vertically downwards so that it reaches the ground in time t0/2??

Answer 1

Do we assume that the boy who throws the boy has some height or that the ball just comes off the ground?

Answer 2

i get an answer of v = 24*t_0
@santistebanc , can you verify that for me

Answer 3

How did you get that answer?

Answer 4

using the height function
\[h(t) = -16t^{2}+v_0t + h_0

sorry equation editor not working for me

Answer 5

hey the options are (a) gt0 (b) 2gt0 (c) 3/4gt0 (d) gt0/4

Answer 6

where g is gravity/acceleration right?

Answer 7

oh they leave it in terms of g

24 = (3/4)*32
--> 3/4g*t_0

Answer 8

yeah that's what I got too. I just use one of the kinematic equaions. The one where you only vi and no vf.

Answer 9

@ dumbcow you are right, the correct answer is C

Answer 10

i dont think I understood the problem

Answer 11

Thanks!)

Answer 12

can anyone give a detailed equation

Answer 13

i cant understand the one which is written i know that answer is correct

Answer 14

Let's wait for dumbcow to answer.

Answer 15

dont you have to consider the difference of height between the two boys?

Answer 16

no

Answer 17

the height function

h(t) = -g/2 t^2 +v_0t +h_0

q throws it up to p such that it reaches highest point with initial velocity v_0 in time t
h_0 = 0
highest point is when velocity = 0
v(t) = -gt +v_0 = 0
--> v_0 = gt
final height is H

--> H = -g/2 t^2 +gt^2
--> H = g/2 t^2

Now set up 2nd throw, which has negative initial velocity and takes time t/2 to reach same height H
initial height is now H
final height is 0

0 = -g/2 (t/2)^2 -v(t/2) +H
substitute in for H

0 = -g/2 (t/2)^2 -v(t/2) + g/2 t^2
solve for v

Answer 18

can some one use kinematics equations to solve this

Answer 19

is the height function not a kinematic equation?
sorry not a physics guy

Answer 20

I assume that all of you know the standard defintions and laws and the variables used
Basically we can divide this problem in 2 parts

Part 1- ball from ground to tower
Let "s" be the height of the tower

First using this equation
\[v = u + at\]

I get
\[0 = u - g t_0\]
\[t_0 = u/g\]

Also
\[s = ut + \frac{1}{2} at^2\]

\[s =\frac{u^2 }{g} -\frac{1}{2} g \frac{u^2}{g^2}\]
\[s =\frac{u^2 }{g} -\frac{1}{2} \frac{u^2}{g}\]
\[s =\frac{1}{2} \frac{u^2}{g}\]

Part 2--> Tower to ground

We know \[s =\frac{1}{2} \frac{u^2}{g}\]

Again using the following equation
\[s = ut + \frac{1}{2} at^2\]
We get

\[\frac{1}{2} \frac{u^2}{g} = u't' + \frac{1}{2}g t'^2\]

Now we know that\[ t ' = \frac{t_0}{2}= \frac{u}{2g}\]

Therefore
\[\frac{1}{2} \frac{u^2}{g} = u'\frac{u}{2g}+ \frac{1}{2}g \frac{u^2}{(2g)^2}\]
\[\frac{1}{2} \frac{u^2}{g} = u'\frac{u}{2g}+ \frac{1}{8} \frac{u^2}{g}\]
\[\frac{3}{8} \frac{u^2}{g} = u'\frac{u}{2g}\]
\[\frac{3}{8} \frac{u}{g} = u'\frac{1}{2g}\]
\[\frac{3}{4} {u} = u'\]

Therefore the boy P should throw the ball down with the velocity which is equal to 3/4 times with which the boy Q threw the ball up

Answer 21

I am sorry if I have given the full answer but it is to dispense all doubts related to the solution. I have reviewed the solution on my side but if you find any mistakes , do notify me

Answer 22

thanxxx

Answer 23

Welcome. I forgot the last step .. Substitute u = gt_0 and you will get your answer