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A bullet is shot horizontally with a speed of 2.0 times 10 to the two meters per second from a height of 1.5 meters.
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How much time elapses before the bullet hits the ground?
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How far does the bullet travel horizontally?
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As we work this exercise, weβll assume that π, the acceleration due to gravity, is exactly 9.8 meters per second squared.
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Weβre told that the bullet is shot with an initial speed of 2.0 times 10 to the two meters per second.
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Letβs call that speed π£ sub π₯, for its speed in the π₯ or horizontal direction.
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The bulletβs initial height is given as 1.5 meters.
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Weβll call that value β.
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The problem involves two parts.
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For the first part, we wanna solve for the time elapsed before the bullet hits the ground.
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Weβll call that time π‘.
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In part two, we wanna solve for the horizontal displacement of the bullet.
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Weβll call that π sub π₯.
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Letβs begin our solution by drawing a diagram of the situation.
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In our scenario, the bullet is fired from the gun with an initial speed π£ sub π₯.
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At first, the bullet is moving only horizontally.
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But then, the force of gravity begins to affect the path of the bulletβs flight.
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Ultimately, the bullet travels a horizontal distance weβve called π sub π₯.
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Before solving for that distance, we want to know how long it takes for the bullet to reach the ground.
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Since the acceleration acting on the bullet is constant, that means we can use then kinematic equations to approach this problem.
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Looking over the equations of motion that are true when acceleration π is constant, we see that the third one from the top can be useful to us to solve for the time π‘.
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Before we put this equation to use, looking back at our diagram, letβs define vertical downward motion as positive and horizontal motion to the right as positive.
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Under this definition, π is equal to positive 9.8 meters per second squared.
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As we apply this kinematic equation to our scenario, we focus on motion exclusively in the vertical or π¦-direction.
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In this direction, π, the distance the bullet travels, is β.
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π£ zero, the initial speed of the bullet in the vertical direction, is zero, so that entire term goes to zero.
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Our acceleration π is equal to π, the acceleration due to gravity, and π‘ is the time it takes for the bullet to fall to the ground.
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So our revised equation says that β, the vertical height for which the bullet drops, is equal to one-half π times π‘ squared.
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Letβs rearrange this equation to solve for π‘.
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Weβll start by multiplying both sides by two divided by π, which cancels out the one-half with the two and both factors of π on the right-hand side.
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If we then take the square root of both sides, that square root cancels out with the squared term leaving us with an equation for π‘ that says: π‘ is equal to the square root of two times β divided by π.
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When we insert the given values for β and the known value for π and enter these values on our calculator, we find a value for the time π‘ of 0.55 seconds.
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Thatβs how long it takes the fire bullet to fall to the ground.
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Now we turn our attention to solving for π sub π₯, the horizontal range of the bullet.
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In the horizontal direction, the bullet moves with a constant speed.
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There is no acceleration.
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When speed is constant, the relationship between speed, distance, and time is given as: speed is equal to distance travelled divided by time.
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In our situation, π£ sub π₯ is equal to π sub π₯ divided by π‘.
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To solve for π sub π₯, letβs multiply both sides by π‘ which cancels that term from the right-hand side.
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So π sub π₯ is equal to π‘ times π£ sub π₯.
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We solved for the value of π‘ previously and weβre given π£ sub π₯ in the problem statement.
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When we insert those two values and multiply them together, we find a distance π sub π₯, to two significant figures, of 110 meters.
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Thatβs the horizontal distance the bullet will travel before it hits the ground.