Video Transcript
A particle was projected vertically upwards from the ground. Given that the maximum height the particle reached was 62.5 meters, find the velocity at which it was projected. Take the acceleration due to gravity 𝑔 equal to 9.8 meters per square second.
In order to answer this question, we will use our equations of motion or SUVAT equations, where 𝑠 is the displacement, 𝑢 is the initial velocity, 𝑣 the final velocity, 𝑎 the acceleration, and 𝑡 is the time. We’re trying to calculate the initial velocity 𝑢. We know that the particle reached a maximum height of 62.5 meters. At the maximum height, the velocity is equal to zero. This means that 𝑣 is equal to zero meters per second. As the acceleration due to gravity is 9.8 and the particle was projected upwards, 𝑎 is equal to negative 9.8 meters per square second.
In this question, we know nothing about the time. The equation we will use is 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. Substituting in our values, we have zero squared is equal to 𝑢 squared plus two multiplied by negative 9.8 multiplied by 62.5. The right-hand side simplifies to 𝑢 squared minus 1225, and this is equal to zero. We can add 1225 to both sides. So, this is equal to 𝑢 squared. Finally, square rooting both sides of this equation gives us a value of 𝑢 of 35.
We can, therefore, conclude that the velocity at which the particle was projected is 35 meters per second.
