atart by fining the initial vertical velocity, and the initial horizontal velocity.
Then, figure the highest point, at which velocity is zero.
Then, note the velocity of the duds. One is zero. At the highest point, the only momentum is due to horizonal velocity, so dud number two must be going now twice the original horizonal velocity.
So find:
a) highest point, and time of hightest point.
b) horizontal distance from launch to hightest point.
c) now find the time for the dud to fall from the hightest point, and then the horizonal distance during falling is d=2vihoriz*timefalling
now add the two horizonal distances, and you can take a break
Suppose a 10.0 kg fireworks shell is shot into the air with an initial velocity of 62.0 m/s at an angle of 80.0 degrees above the horizontal. At the highest point of its trajectory, a small explosive charge separates it into two pieces, neither of which ignite (two duds). One 9.00 kg piece falls straight down, having zero velocity just after the explosion.
(a) At what horizontal distance from the starting point does the 9.00 kg piece hit the ground?
(b) Calculate the velocity of the 1.00 kg piece just after the separation.
(c) At what horizontal distance from the starting point does the 1.00 kg piece hit the ground?
**** I apologize in advance for not having any work to show, but its because I have no clue on how to approach this problem. If I could get some help on what equations to use, that would help.
1 answer
1 answer