![]() ![]() ![]() To do this, we can look at a diagram of the initial speed of the projectile and the angle that it was launched above the horizontal. All that’s left for us to do is to find an expression for the initial vertical velocity of the projectile □ □. Tidying this up slightly, we get □ equals two □ □ divided by □, and this is equal to our time of flight capital □. Finally, we’ll divide both sides of the equation by □, and these □’s on the left will cancel each other out. Then we’ll multiply both sides by two, where these twos on the left cancel each other out. We’ll first add a half □□ to both sides, and these terms on the right cancel each other out. So we want to solve the following equation: zero equals □ □ minus a half □□. The second time the vertical displacement of the projectile is zero is at the end of the motion. At the beginning of the motion when □ equals zero, the vertical displacement of the projectile is also zero. This expression will equal zero at two values of □: one when □ equals zero and the other when the bit inside the brackets, □ □ minus a half □□, is equal to zero. Then we can take □ as a common factor, giving zero equals □ □ minus a half □□ all multiplied by □. The point in the projectile’s motion that we’re interested in is when the vertical displacement of the projectile is equal to zero, so we can write zero equals □ □ multiplied by □ minus a half □□ squared. Notice that there is a negative sign here because gravity acts downwards as highlighted on our diagram. So we can write □ □ is equal to the projectile’s initial vertical velocity □ □ multiplied by time minus one-half of □ multiplied by time squared. In our case, we want to look at the vertical displacement of the projectile, which we will call □ □. We will start with the following formula, which tells us that the displacement of an object undergoing constant acceleration is equal to its initial speed multiplied by time plus one-half of the acceleration it is experiencing multiplied by time squared. ![]() So we need an equation that relates displacement to the time during the projectile motion. To answer this question, we need to find the time at which the projectile reaches a certain displacement. The question asks us to work out the time at which it returns to the ground again, which we will call capital □, and is also known as the time of flight of the projectile. ![]() And at the end of the motion, its vertical displacement is equal to zero. We will call the vertical displacement of the projectile □ □. And we are told that the projectile returns to the ground at the same height that it was launched from. This downward force causes the projectile’s path to be curved. During projectile motion, there is only the downward force from gravity acting on the projectile, which has a magnitude of the mass of the projectile, which we will call □, multiplied by □, the acceleration due to gravity. We have a projectile launched with an initial speed that we will call □, and it’s launched at an angle above the horizontal that we will call □. What is the time between the projectile leaving the ground and returning to the ground at the same height that it was launched from? A projectile has an initial speed of 25 meters per second and is fired at an angle of 48 degrees above the horizontal. ![]()
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