A 4.0 kg shell is shot with an initial velocity of magnitude 20.0 m/s at an angl

Question

A 4.0 kg shell is shot with an initial velocity of magnitude 20.0 m/s at an angle of 60o with the horizontal. At the top of the trajectory, the shell explodes into two fragments. (a) (4 points) Find how long it takes for the shell to rise to the top of its trajectory. t = ___________________ (b) (4 points) Find the velocity (magnitude and direction) of the shell just before the explosion. V = ___________________ Just after the explosion, the velocity of the 1.0 kg fragment #1 is 1.0 m/s horizontally. (c) (4 points) Find the velocity (magnitude and direction) of fragment #2 just after the explosion. V2 = __________________ (d) (4 points) Find the distance from the gun to the point where fragment #2 lands, assuming the terrain is level and that air drag is negligible. D = ________________ (e) (4 points) Which fragment hits the ground first? Fragment # ____________

Explanation / Answer

T=usin/g=1.732s,Hmax=u2(sin)2/2g=30m,R=u2sin2/g=34.64m

vh=ucos=10 ms-1 direction horizontal

using conservation of linear momentum in the horizontal

4.10=1.1+3.v2.

v2=13 ms-1 direction horizontal

after explosion both have zero initial velocity in vertical so both will fall at the same time(t)

Hmax=0.5gt2

t=2.45s

Distance to which the particle 2 lands=34.64/2+2.45*13=49.163 m

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