A projectile returns to its original height 4.08 s after being launched, during which time it travels 76.2 m horizontally.If air resistance can be neglected, what was the projectile's initial speed?Express your answer in m/s.

To solve this problem, we need to break it down into two parts: the horizontal and vertical components of the projectile's motion.

1. Horizontal motion: The horizontal distance (d) covered by the projectile is given as 76.2 m. The time (t) taken is 4.08 s. We know that distance = speed * time. Therefore, the horizontal speed (Vx) of the projectile can be calculated as:

   Vx = d / t Vx = 76.2 m / 4.08 s Vx = 18.676 m/s

2. Vertical motion: The time taken for the projectile to return to its original height is 4.08 s. This means that the time taken to reach the maximum height is half of this, which is 2.04 s. We know that the vertical speed (Vy) can be calculated using the equation of motion: Vy = g * t, where g is the acceleration due to gravity (approximately 9.8 m/s²). Therefore:

   Vy = g * t Vy = 9.8 m/s² * 2.04 s Vy = 19.992 m/s

3. The initial speed of the projectile is the vector sum of the horizontal and vertical speeds. This can be calculated using Pythagoras' theorem: V = sqrt(Vx² + Vy²). Therefore:

   V = sqrt((18.676 m/s)² + (19.992 m/s)²) V = sqrt(348.9 m²/s² + 399.68 m²/s²) V = sqrt(748.58 m²/s²) V = 27.36 m/s

So, the initial speed of the projectile was approximately 27.36 m/s.