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kamran8 Group Title

A projectile is fired at an upward angle of 45 from the top of a 256m cliff with a speed of 185 m=sec. What will be its speed when it strikes the ground below?

  • 10 months ago
  • 10 months ago

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  1. DemolisionWolf Group Title
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    |dw:1382126196449:dw| we will break this question up into x and y components of speed (aka velocity), and into 3 events: event1 - when projectile is fired, event2-when projectile reaches it's peak, and event 3- when the projectile strikes the ground

    • 10 months ago
  2. DemolisionWolf Group Title
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    here are the events, and what we know about each one of them. next we use the kinematic equations to solve for values that will get us from event 1 to event2, then from event 2 to event 3, so we can find out the speed the projectile hits the ground at. |dw:1382126610154:dw|

    • 10 months ago
  3. DemolisionWolf Group Title
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    The kinematic equations are: V + at = V V^2 + 2aS = V^2 S+vt+(1/2)at^2 = S these 3 equations are the 'math' way of saying basically this: event 1 = event 2, so lets take the actual kinematic equation above and see how it is like the 'event' equation I just told you event 1 = event 2 v + at = v can you see how the "v + at" is on the same side as the 'event 1' part? and how 'v' on the other side of the = sign lines up on the same side as the event 2 part? So, what we are saying is that these variables from event one, when added and multiplied this way will equal the velocity of event 2. so we will use the "_1" to refer to event 1 and "_2" to refer to event 2. now our kinematic equation looks like this: v_1 + a_1 * t_1 = v_2 now if I were to describe the equation I would say, "velocity at event 1 plus acceleration at event 1 times time at event 1 equals velocity at event 2" we need to add one more symbol onto our "_1" and "_2" symbols, it is direction of x and y. These equations only work when we use the numbers of a movement in the same direction. so if event 1 values were all movements in the y direction, then event 2 will be a value of movement in the y direction also. we will use the notation of "_1y" and "_2y" which now will mean, 'event 1 in the y direction' and the other will mean 'event 2 in the y direction'

    • 10 months ago
  4. DemolisionWolf Group Title
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    lets begin: first figure out time it takes to reach the apex of the projectile's path- [event1 = event2] V + at = V (V_1y) + (a_1y)*(t_1y) = V_2y sin(45)*185 + (-9.81)*(t_1y) = 0 [this =0 because at the apex, velocity = 0] t_1y = 13.33 sec Solve for height the projectile reaches: [event 1 = event2] S+vt+(1/2)at^2 = S (S_1y) + (v_1y)*(t_1y) + (1/2)*(a_1y)*(t_1y)^2 = S_2y 265 + sin(45)185 * 13.33 + (1/2) (-9.81) (13.33^2) = S_2y S_2y = 1137.2 m Solve for velocity in the y direction when projectile hits the ground [event 2 = event 3] V^2 + 2aS = V^2 (V_2y)^2 + 2*(a_2y)*(S_2y) = (V_3y)^2 0 + 2*(-9.81)*(1137.2) = (V_3y)^2 V_3y = 149.3 m/s combine x and y components of velocity at ground (see picture below) [event 3] let, V_1x = V_2x = V_3x then if a^2 + b^2 = c^2 let a = V_3x, b = V_3y, c = V_3, velocity at impact [answer] (V_3x)^2 + (V_3y)^2 = V_3 (cos(45)*185)^2 + (149.3)^2 = V_3 |dw:1382129041312:dw|

    • 10 months ago
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