In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a shelf above the point where the quarter leaves your hand and is a horizontal distance of 2.1 m from this point (Fig. E3.19). If you toss the coin with a velocity of 6.4 m/s at an angle of 60° above the horizontal, the coin will land in the dish. Ignore air resistance. (a) What is the height of the shelf above the point where the quarter leaves your hand?

Video transcript

Hey everyone. So today we're being asked to solve a problem about projectile motion. So we're given a scenario in which we are with some friends and we are throwing a disc on top of the mound At an angle of 35° to the horizontal. It's an upward angle, it's above the horizontal and the mound is 36 m away. Given that the initial velocity, the launching speed of the ball is 24 m/s. We're being asked to find the height of the rocky mount. So before going into any other details, let's just draw sort of schematic if you will. So if we have a horizontal plane here, just a straight horizontal and yes, the projectile is landing at a higher point in this case this is the change in the y. But for any projectile motion, let us say that this is the direction of the projectile. If it's being launched at an angle theta above the horizontal, then that means not only is there a horizontal aspect to the motion, the horizontal part, there is also the vertical part of the motion. The vertical component. Let me write that a little clearer. That is an A. So we have everything we need for the horizontal component. And this is because for the horizontal component of velocity and projectile motion, the only formula we can use that the change in horizontal distance is equal to the initial velocity in the X direction, multiplied by time. And the reason we don't have uniformly accelerated motion equation, quote unquote. For the horizontal motions that in projectile motion, the velocity velocity in the or the initial velocity in the X direction is constant is constant and if velocity doesn't change then that means the acceleration is equal to zero. In a similar vein in the vertical aspect, in the vertical component of the velocity, the acceleration is equal to the negative force of gravity. And this is because we are acting against gravity and then gravity is again pulling the projectile down once the motion starts. But with this in mind, the last thing we need to take note of is that to find this initial velocity in the X direction and initial velocity in the Y direction even we actually need to use a bit of trigonometry because the launching speed given here, the initial velocity is a component. It has both the x and Y. Parts included in it. We're trying to separate that out and the way we do that is by utilizing some basic trigonometry. So if we're, I've drawn this out as a right triangle, even though projectile motion follows a more parabolic path because it's very easy to visualize the trig aspect here. If we have the angle data here, then this horizontal component, This horizontal component right here lies adjacent, it is adjacent right and the vertical component lies opposite to the angle it is directly across is not touching the angle on either sides. Like the hypotenuse or the horizontal aspect with this, we can go ahead and write out our trick identities. We have so car oops. Alright this little to the side. Oops. Our trig identities are so tower. So if it's adjacent, if the side in question is adjacent, we use Kassian to help figure out its value and if it's opposite we use sign to figure out its value. So by that logic, the vertical aspect or sorry, the horizontal aspect is equal to the initial velocity. The component velocity multiplied by cosine theta because again the horizontal aspect is adjacent and the vertical aspect, the vertical is equal to uh the initial component velocity multiplied by sine data. And let's just solve these out. This will be 24 m/s, multiplied by Kassian degrees. Excuse me. Which will be 19.66 m/s. Similarly 24 m per second multiplied by sine 35 degrees. And that says the value of 13. m/s. We're going to put a pin in the vertical velocity for now. We're going to need it later. But back to our initial horizontal aspect because to solve for the vertical height, we need the time that the projectile was in the air that it was in motion. And that's why we're using the horizontal aspect to help us find that. So we know that the distance traveled was 36 m away Right here Horizontally. So it'll be 36 m is equal to 19.66 m/s. What we calculated over here multiplied by time and I'll write that in black By time. So therefore time is equal to 36 m divided by 19.66 meters per second, which equals 1.83 seconds 83 seconds. Now we can use the formula, the uniformly accelerated motion formula that the change in distance, our change in vertical height is equal to the initial velocity in the Y direction, multiplied by time plus one half acceleration, times times square acceleration in the y direction, of course was applied by times square. So substituting in our values because we actually have all of them, We get that this is equal to 13.77 m/s Multiplied by our newly acquired time, which is 1.83 seconds plus one half. You have 1/2 in two -9.8 again because it's the negative value of G in the vertical direction meters per second squared Multiplied by and I run out of space. So I'll just put it down here, 1.83 seconds squared. That's outside the parentheses, of course, simplifying this, We get that the change in height or the height of the mound is equal to 8. m. Remember if we cross out our units, we have second squared here, second squared here and we are good to go. So with that the height of the mound is answer choice D 8.79 m. I hope this helps, and I look forward to seeing you all in the next one.