A projectile was fired vertically upwards from the Earth’s surface at 1,218 meters per second. It hit a target that was 1,575 meters above the surface of the Earth. Find the speed of the projectile when it hit the target. Consider the acceleration due to gravity to be 9.8 meters per square second.
We are told that a projectile is fired vertically upwards from the Earth’s surface. We know that the initial speed is 1,218 meters per second. The target is 1,575 meters above the surface. We are asked to calculate the speed when the projectile hits the target. We are also told that the acceleration due to gravity is 9.8 meters per square second. This will be working towards the Earth, which means that the velocity of the particle will be decreasing as the projectile is decelerating.
We can answer this question using the equations of uniform acceleration, also known as the SUVAT equations. Our values of 𝑠, 𝑢, and 𝑎 are 1,575, 1,218, and negative 9.8, respectively. We want to calculate the value of 𝑣. In order to do so, we will use the equation 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. Substituting in our values, we have 𝑣 squared is equal to 1,218 squared plus two multiplied by negative 9.8 multiplied by 1,575. Typing the right-hand side into the calculator gives us 1,452,654.
We can then take the square root of both sides of this equation so that 𝑣 is equal to 63 root 366. This is the velocity of the projectile when it hits its target. We could also write this answer as a decimal. The velocity is equal to 1,205.3 meters per second to one decimal place.