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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?

Engineering
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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
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|
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'

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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|

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