A ball is dropped from a height of 100 m at t = 0. Later, at t = 1.00 s, a second ball is thrown downward with a speed of 19.8 m/s. At what time will the two balls be at the same height?

2 Answers
Jane
Jan 28, 2018

#88.6 m# above the ground after #1.510 s# of the 1st ball was thrown (#g=10 m/s^2# is taken)

Explanation:

Suppose,after falling through distance #x# from the point of release,both the balls will meet.

Now,for reaching this distance,if the 1st ball takes time,#t#,

Then, #x= 1/2 g*t^2#...#1#

And,for the 2nd ball if #t'# time is required,then,

#x= 19.8t'+1/2g*t'^2#....#2#

Now,given, #t-t'=1# (as #1 s# after the release #1st# ball,the #2nd# ball was thrown)

So,putting the value of #t=1+t'# in equation #1# we can solve the equations #1# &#2# and we get, #t'=0.510 s#

#And, t=1.510 s#

So,put the value in equation #2# and we get, #x= 11.4 m#

And that means #(100-11.4) m# or #88.6 m# above the ground

MattyMatty
Jan 28, 2018

#1.49 " seconds"#

Explanation:

#t = sqrt(2*h/g)#
#=> h_1 = t^2*g/2#
#v = v_0 + g t#
#h_2 = 19.8 * (t - 1.00) + g * (t - 1.00)^2 / 2, t >= 1#

#"Here we have :"#
#h_1 = h_2#
#=> t^2*g/2 = 19.8*t - 19.8 + g*t^2/2 - g*t + g/2#
#=> 19.8*t - 19.8 - g*t + g/2 = 0#
#=> (19.8 - g)*t = 19.8 - g/2#
#=> t = (19.8 - g/2)/(19.8 - g)#
#= (19.8 - 4.9)/(19.8 - 9.8)#
#= 14.9/10#
#= 1.49 s#

#g "= gravity constant = 9.8 m/s²"#

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