Ignoring air resistance,a catapult on a level field tosses a 5.0-kg stone a horizontal distance of 43 m.A second 5.0-kg stone tossed in an identical fashion breaks apart in the air into 2 pieces,one with a mass of 3.2 kg and one with a mass of 1.8 kg.The breaking apart happens in such a way that both of the pieces hit the ground at the same time.If the 1.8-kg piece lands a distance of 59 m away from the catapult,how far away from the catapult does the 3.2-kg piece land?
A)15 m
B)27 m
C)34 m
D)50 m
E)52 m

Question

Quizplus · Accepted Answer

Since both pieces hit the ground at the same time, we can use the equation for horizontal distance: 
d = (v^2*sin(2θ))/g 
where d is the horizontal distance, v is the initial velocity, θ is the launch angle, and g is acceleration due to gravity. 
Since the initial conditions are identical for both stones, we can set up a ratio: 
(3.2/5) = (d/59) 
Solving for d, we get: 
d = 37.12 m 
Therefore, the 3.2-kg stone lands 43 - 37.12 = 5.88 m away from the catapult.

Ignoring Air Resistance,a Catapult on a Level Field Tosses a 5.0-Kg