A projectile is launched with an initial speed . At its highest point, the projectile's speed is . What was the launch angle above the horizontal?
(in deg)
Not enough INFO.
To find the launch angle above the horizontal, we can use the concept of projectile motion.
In projectile motion, the vertical and horizontal components of motion are independent of each other. The initial vertical velocity of the projectile determines the maximum height reached, while the horizontal velocity remains constant throughout the motion.
Let's assume that the initial speed of the projectile is v₀ and the speed at its highest point is v_max.
At the highest point, the vertical velocity is zero because the object momentarily stops moving upward before falling back down. Therefore, we can equate the vertical component of the launch velocity to zero at this point:
v_y = v₀ * sin(θ) = 0
This equation tells us that the launch angle θ must be zero or 180 degrees. However, a launch angle of 180 degrees would mean the object is launched straight downward, which is not possible since we are considering a projectile launched above the horizontal. Hence, the launch angle must be zero degrees.
Therefore, the launch angle above the horizontal is 0 degrees.