Question #22257

an object is thrown upwards from a height of 40.0 m. A second object is dropped from that height 3.0 s later. What must be the initial speed of the first ball if both are to hit the ground at the same time?

Expert's answer

Let y1(t) and y2(t) be the heights of the first and thesecond balls at moment t respectively.

Let also v be the initial velocity of the first ball, andh1=40m be the initial height.

Then v1 is directed up.

Since both ball move with constant accelerationg=9.8m/s^2 directed down, we have that

y1(t) = h1 + v*t- g*t^2/2

y2(t) = h1, for t < 3s

= h1 -g*(t-3)^2 / 2, for t > 3s

Let s be the moment at which both balls hit the ground.

Then

y1(T) = y2(T) =0.

Hence from

y2(T) = 0

we get

h1 - g*(T-3)^2/ 2 = 0

g*(T-3)^2 / 2 =h1

T = 3 +square_root(2*h1/g)

= 3 +square_root(2*40/9.8)

= 5.8571seconds

Substitute the value of s into the identity:

y1(T) = 0

h1 + v*T -g*T^2/2 = 0

v*T = g*T^2/2 -h1

v = g*T/2 -h1/T

=9.8*5.8571/2 - 40/5.8571

= 21.870 m/s

Answer: v = 21.870 m/s

Let also v be the initial velocity of the first ball, andh1=40m be the initial height.

Then v1 is directed up.

Since both ball move with constant accelerationg=9.8m/s^2 directed down, we have that

y1(t) = h1 + v*t- g*t^2/2

y2(t) = h1, for t < 3s

= h1 -g*(t-3)^2 / 2, for t > 3s

Let s be the moment at which both balls hit the ground.

Then

y1(T) = y2(T) =0.

Hence from

y2(T) = 0

we get

h1 - g*(T-3)^2/ 2 = 0

g*(T-3)^2 / 2 =h1

T = 3 +square_root(2*h1/g)

= 3 +square_root(2*40/9.8)

= 5.8571seconds

Substitute the value of s into the identity:

y1(T) = 0

h1 + v*T -g*T^2/2 = 0

v*T = g*T^2/2 -h1

v = g*T/2 -h1/T

=9.8*5.8571/2 - 40/5.8571

= 21.870 m/s

Answer: v = 21.870 m/s

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