A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.030.0^\circ above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and distances from where the projectile was launched to where it lands?

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Final Answer

x=130 mx = 130 \textrm{ m}, y=30.9 my = 30.9 \textrm{ m}

Solution video

OpenStax College Physics, Chapter 3, Problem 25 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. Because this projectile does not land at the same height that it was launched from, we cannot use this range formula. This range formula assumes the projectile lands at the same vertical position of the launch. So, we have to examine each of the x and y components separately and then figure out what the x position and the y position of the target is going to be. So the projectile is launched at an angle of 30 degrees above the horizontal at the speed of 50.0 meters per second. We know that there is no horizontal acceleration, so a subscript x is zero. There is a vertical acceleration due to gravity of negative 9.8 meters per second squared. So this acceleration due to gravity will have an effect on the vertical component of the velocity, and we're going to take the vertical position to be y naught equals zero and x naught equals zero and draw [inaudible 01:07] to figure out what is x and what is y of this position three seconds after the launch. So the x position will be the initial x position plus the initial x component of the velocity times time, and this is constant so there's no acceleration term here at all because there is no acceleration in the x direction. So we have zero for the initial x position, and so this becomes v naught times cosine of theta times t. It's v naught which is this vector in green, initial velocity times cosine of the angle theta because vx component is the adjacent leg of this right triangle. Then we multiply that by the time of three seconds. So that's 50 meters per second times cos 30 giving us the x component of the velocity, multiplied by three seconds means, it will move horizontally 130 meters. Then for the y component we say that it will be its initial y position, which is zero, plus the y component of the velocity initially, multiplied by time, plus this acceleration term which is present because the acceleration in the y direction is negative 9.8, it's non-zero. So we have v naught time sine theta because the y component of this velocity is the opposite leg of the triangle and so the trigonometric function sine is what we'll use to get the opposite leg, multiply it by the hypotenuse. So we have v naught sine theta times the time, plus one half times the vertical component of the acceleration which is the acceleration due to gravity, times time squared. So that's 50 meters per second times sine 30, times three seconds, plus a half times negative 9.8 meters per second squared, times three seconds squared, gives a vertical position of 30.9 meters above the height of the launch.


Hi iliketurtles, thank you for the question. a_x is the horizontal acceleration, and it is zero since we're neglecting air drag. Without air friction there is nothing to apply a horizontal force. Gravity is entirely vertical, which is why a_y is not zero. Gravity has no horizontal component since we've chosen our coordinate system such that gravity is only vertical.
Hope this helps,

ax = 0 as there is no horizontal acceleration. There is velocity but it is a constant.
The vertical acceleration is do to gravity but this does not exist on a horizontal plane, only the vertical plane.

Where does the sign and cosign of theta come from in the two equations?

Hi Chrousis, we need to consider components of the velocity along the 'x' and 'y' directions. Sine of the angle times the velocity gives the vertical component (the 'y' direction in other words), whereas multiply the velocity by cosine of the angle gives the horizontal component.
All the best with your studies,