Question

a bullet has a speed of 350 m/s as it leaves a rifle it is fired horizontally from a cliff 6.4m above a lake how far does the bullet travel before striking the water

Answer 1

1) A bullet has a speed of 350 m/sec as it leaves a rifle. If it is fired horizontally from a cliff 6.4 m above a lake, how far does the bullet travel before striking the water?Solution:We have a 2-dimensional problem with constant acceleration (acceleration due to gravity). This is a projectilemotionproblem. The figure is as shown and the coordinate system selected is drawn. The origin is placed at the bullet's location at time t=0. Hence the initial conditions for the problem are:x(t=0) = xo= 0 ; y(t=0) = yo= 0vx(t=0) = vox= 350 m/s; vy(t=0) = voy= 0xy6.4 mvoaSince the only force acting is gravity (downward = + y direction), we have: ax= 0; ay= + g = + 9.8 m/sec2. The general solutionsfor the constant acceleration problem in two dimensions are:x(t) = (1/2) axt2+ voxt + xoy(t) = (1/2) ayt2+ voyt + yovx(t) = axt + voxvy(t) = ayt + voyInserting the values of acceleration and the initial conditions gives us the specific equations(applicable to this one particular problem).x(t) = (350) t y(t) = (1/2)(9.8) t2vx= 350 m/s vy(t) = 9.8 tLet t' be the time when the bullet hits the lake. We then know that: y(t') = + 6.4 m. Thus:y(t') = + 6.4 = + 4.9 t'2t' = 1.143 sec.The horizontal (x) position of the bullet at this time is then: x(t') = (350)(1.143) = 400 m.2) A player kicks a football at an angle of 37owith the horizontal and with an initial speed of 48 ft/sec. A second player standing at a distance of 100 ft from the first in the direction of the kick starts running to meet the ball at the instant it is kicked. How fast must he run in order to catch the ball before it hits the ground?Solution:We have a projectile motion problem (as far as the football is concerned). Hence we have drawn a figure, chosen a CS, and write down the initial conditions (initial position & velocity) of the football (at t=0).x0= 0; y0= 0; v0x= v0cos 37; v0y= v0sin 37The acceleration is: ax= 0; ay= - 32 ft/sec2.xy37voa03-2The general equationsof motion for constant acceleration in 2-dimensions are:

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x(t) = (1/2) axt2+ voxt + xoy(t) = (1/2) ayt2+ voxt + yovx(t) = axt + voxvy(t) = ayt + voyWe insert the known values for acceleration & initial conditions and obtain the specific equationsfor the football:x(t) = (48)(4/5) t

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