![]() The speed is the square root of the sum of the squares of each of the horizontal and vertical components for velocity. That’s 49 over two minus 9.8 times three, which is negative 49 tenths. But we do know that □ is equal to three. We’re looking to find the speed three seconds after it was thrown. So, □ □, the velocity at any point, is 282 over five. Now, in fact, we don’t need to do this in the horizontal direction since we said acceleration is zero. This time, we use the equation □ equals □ plus □□. So, we’ll work out the horizontal and vertical velocity and then find the magnitude. Now, remember, speed is the magnitude of the velocity. Now that we have all this information, we can work out the speed three seconds after the projectile was thrown. And we find □ sub □ to be equal to 49 over two. We’re going to add 245 over two to both sides to get 245 over two equals five □ sub □. And this equation simplifies to zero equals five □ sub □ minus 245 over two. ![]() Then, a half □□ squared is a half times negative 9.8 times five squared. And once again, we know that this takes five seconds. It ends up at exactly the same height as it started. But we do know that the vertical displacement is zero. We don’t know the velocity at a given point, although we could potentially work it out. Acceleration due to gravity acts in the opposite direction as negative 9.8. We defined the initial velocity in the vertical direction to be □ □. Let’s repeat this process for the vertical motion. And we get □ sub □ to be equal to 282 over five or 282 over five metres per second. And we can solve this equation by dividing through by five. So, we have 282 equals □ sub □ times five. And then, a half □□ squared is a half times zero times five squared. Horizontally, the total displacement is 282. And the reason behind using this one in particular will become evident in a moment. We’re going to use □ equals □□ plus a half □□ squared. We’re going to use one of the equations of constant acceleration to find the initial velocity of the particle. It travels a total distance of 282 metres and it takes five seconds to do so. So its velocity at any point is still □ sub □. And this means its horizontal velocity remains unchanged. We know that the acceleration horizontally is zero. Our aim is to find the initial speed in both directions. We’re now going to consider the motion of particle in both the horizontal and vertical direction. And vertically, its initial speed is □ sub □ upwards. So horizontally, its initial speed is □ sub □. So, what we’re going to do is split it into the horizontal and vertical components of the initial speed. Acceleration due to gravity acts in the vertical direction that’s downwards. It covers a total distance of 282 metres horizontally. The motion of the particle looks a little something like this: it’s an inverted parabola. Let’s begin by drawing a sketch of what’s going on here. Take □ to be equal to 9.8 metres per square second. ![]() Find, to one decimal place, a projectile’s speed three seconds after it was thrown, given that it flew a total horizontal distance of 282 metres for five seconds before hitting the ground.
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