Question #153020

A bullet is fired vertically into the air from the top of building and reaches a height of 1600m. It then falls to the ground below.The total time of flight is 38.0 s.


i) Determine the speed at which the bullet leaves the gun


ii) Determine from what height above the ground is the bullet fired


Determine the speed of the bullet when it hits the ground


Sketch a velocity versus time graph for the motion of the bullet



Expert's answer

i) mgh=mv02/2→v0=2gh=2⋅9.8⋅1600=177(m/s)mgh=mv_0^2/2\to v_0=\sqrt{2gh}=\sqrt{2\cdot9.8\cdot1600}=177(m/s) . Answer


ii) v=v0−gt→t=v0/g=177/9.8=18.1(s)v=v_0-gt\to t=v_0/g=177/9.8=18.1(s)

H=gt2/2=9.8⋅(38−18.1)2/2=1940.4(m)H=gt^2/2=9.8\cdot(38-18.1)^2/2=1940.4(m)

h0=1940.4−1600=340.4(m)h_0=1940.4-1600=340.4(m) . Answer


iii) mgH=mv2/2→v=2⋅9.8⋅1940.4=195(m/s)mgH=mv^2/2\to v=\sqrt{2\cdot9.8\cdot1940.4}=195(m/s) . Answer


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