Answer:
Time = 0.55 s
Height = 8.3 m
Explanation:
The ball is dropped and therefore has an initial velocity of 0. Its acceleration, g, is directed downward in the same direction as its displacement, [tex]h_b[/tex].
The dart is thrown up in which case acceleration, g, acts downward in an opposite direction to its displacement, [tex]h_d[/tex]. Both collide after travelling for a time period, t. Let the height of the dart from the ground at collision be [tex]h_d[/tex] and the distance travelled by the ball measured from the top be [tex]h_b[/tex].
It follows that [tex]h_d+h_b=9.8[/tex].
Applying the equation of motion to each body (h = v_0t + 0.5at^2),
Ball:
[tex]h_b=0\times t
+ 0.5\times 9.8t^2[/tex] (since [tex]v_{b0} =0[/tex].)
[tex]h_b=4.9t^2[/tex]
Dart:
[tex]h_d=17.8\times t - 0.5\times9.8t^2[/tex] (the acceleration is opposite to the displacement, hence the negative sign)
[tex]h_d=17.8\times t - 4.9t^2[/tex]
But
[tex]h_b+h_d =9.8[/tex]
[tex]17.8\times t - 4.9t^2+4.9t^2 =9.8[/tex]
[tex]17.8\times t = 9.8[/tex]
[tex]t = 0.55[/tex]
The height of the collision is the height of the dart above the ground, [tex]h_d[/tex].
[tex]h_d=17.8\times t - 4.9t^2[/tex]
[tex]h_d=17.8\times 0.55 - 4.9\times(0.55)^2[/tex]
[tex]h_d=9.79 - 1.48225[/tex]
[tex]h_d=8.3[/tex]
