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?
1
Expert's answer
2013-01-21T10:31:22-0500
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
Comments
Leave a comment