A projectile is caught at the same height from which it was launched. If it is caught with a velocity of -15m/s, what is the initial velocity?
To determine the initial velocity of the projectile, we will use the kinematic equation that relates the final velocity (vf), initial velocity (vi), acceleration (a), and displacement (d). In this scenario, the displacement is zero since the projectile is caught at the same height from which it was launched.
The kinematic equation we'll use is:
vf^2 = vi^2 + 2ad
Since the displacement is zero, the equation simplifies to:
vf^2 = vi^2
Now, let's plug in the given values:
vf = -15m/s (the negative sign indicates the opposite direction from the launch)
vi = ? (initial velocity)
Substituting the values into the equation, we get:
(-15m/s)^2 = vi^2
Solving for vi:
225 = vi^2
To find the value of vi, we take the square root of both sides of the equation:
√225 = √(vi^2)
This gives us:
15 = |vi|
Since speed is always a positive quantity, we take the absolute value to remove the negative sign:
vi = 15m/s
Therefore, the initial velocity of the projectile is 15 m/s.