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First question is..

A shot is fired horizontally at a target 20m away and it is notice3d that the bullet hits a point 7.5cm below the target aimed at. What was the velocity of the bullet when it left the pistol?

I get vertically

u=0ms^1 a=9.8ms^2 s=0.075m need t

so i used s=ut+0.5at^2

=0.075m=4.9t^2

t^2=0.075/4.9ms^2 so t =sqrt of 0.015 so t = 0.122seconds

i the put this value int0 s=ut+0.5at^2

so horizontally 20m=ux0.122+0.5(9.8ms^2x0.122^2)

=20=ux0.122+0.073 so 19.937/0.122 =u so u = 163ms^1

so the velocity on leaving the pistol was 163ms^1

second question.

A shot is fired with a velocity of 100ms^1 at an angle of 25degrees above a horizontal plane.Find...a) the time of flight on the horizontal plane b) the range on the horizontal plane. c)the greatest heiht the shot attains above the horizontal plane. d)the velocity of the shot (magnitude and direction)8.0seconds after it was fired.

for a i get 8.73 seconds b) i get 790metres c)i get 91m and d) i get 120ms^1 at 25degree to the hoizontal.

i am having a little bit of a job getting my head around this topic so if thes are not correct can someone help me out with what im doing wrong?

Kind regards,

Chris.