To
solve this problem, we make use of the equations for linear motion. The relevant
formula to use here is:
<span>y
= vi t + 0.5 a t^2 --->
1</span>
where,
<span> t = time, y = distance, a = acceleration =
gravity, vi = initial velocity</span>
Let
us say that the light rock is 1, and the heavy rock is 2. We know that the
distance of the two rocks must be equal, therefore:
y1
= y2
vi1
t1 + 0.5 g t1^2 = vi2 t2 + 0.5 g t2^2
From
the given values, we know that rock 1 is simply dropped therefore vi1 = 0,
therefore:
<span>0.5
g t1^2 = vi2 t2 + 0.5 g t2^2 --->
2</span>
We
also know that t1 = t2 + 0.5 or t2 = t1 – 0.5. Therefore we need first to find
for the value of t1. By using equation 1:
55
= 0.5 (9.8) t1^2
t1^2
= 11.22
t1
= 3.35 s
Therefore:
t2
= t1 – 0.5 = 3.35 – 0.5
t2
= 2.85 s
<span> Going back to equation 2:</span>
0.5
(9.8) (3.35)^2 = vi2 (2.85) + 0.5 (9.8) (2.85)^2
2.85
vi2 = 15.19
vi2
= 5.33 m/s
<span>Therefore
he must throw the rock at an initial velocity of 5.33 m/s to reach the 1st
rock.</span>
