PLEASE HELP WITH THIS PHYSICS QUESTION!!??

A ball thrown vertically upward with an initial speed of 2.8 from a window that is 3.6m above the ground. How long after the first ball is thrown should a second ball be simply dropped from the same window so that both balls hit the ground at the same?

2 Answers
Mar 6, 2017

Answer:

#t_2 = 0.332#[s]

Explanation:

The first ball describes the path given by

#y_1 = h_0 + v_1 t_1 - 1/2 g t_1^2#

At time #t_2# the second ball is released so

#y_2 = h_0 -1/2g(t_2-t_1)^2#

and both hit the ground at the same time.

#{(h_0 + v_1 t_1 - 1/2 g t_1^2=0),(h_0 -1/2g(t_2-t_1)^2=0):}#

Solving for #t_1, t_2# we obtain a set of four solutions. The feasible one is

#t_1 = (v + sqrt[2 g h + v^2])/g, t_2 = (-sqrt[2] sqrt[g^3 h] + g (v + sqrt[2 g h + v^2]))/g^2# giving

#t_1 = 1.188# when both balls hit the ground and #t_2 = 0.332# the time when the second ball is released.

Attached a plot showing both paths.

enter image source here

Mar 8, 2017

Answer:

#0.33s#, rounded to two decimal place

Explanation:

Time taken by the second ball to hit ground after it is just dropped is governed by the kinematic equation
#h=ut+1/2g t^2# ........(1)
Inserting given values and taking #g=9.8ms^-2#, we get
#3.6=0xxt+1/2xx9.8 t_2^2#
#=> t_2^2=3.6/4.9#
Ignoring the negative root as time can not be negative we get
#=> t_2=6/7=0.857s#

Time taken by the first ball to hit ground after it is thrown vertically up is also governed by the kinematic equation (1). However, in this case gravity is acting against the direction of motion and distance is also negative as ball hits the ground. We get
#-3.6=2.8t_1+1/2xx(-9.8) t_1^2#
#=>49 t_1^2-28t_1-36=0#
This quadratic equation can be solved using the formula
#x=(-b+-sqrt(b^2-4ac))/(2a)#

I used inbuilt graphics tool to find out the roots. Valid root for time is
my comp
#t_1=1.189s#
If both balls are to hit ground together, second ball should be released after time
#t_1-t_2#
#=1.189-0.857#
#=0.33s#, rounded to two decimal place