Question

Given the initial velocity components of a ball as:

Vy = 19.6 m/s
Vx = 33.9 m/s
At a launch angle of 30 degrees, which answer best gives the horizontal distance the ball travels before returning to its starting height?

Answer 1

The horizontal distance the ball travels before returning to its starting height is approximately 135.6 meters.

Step-by-step explanation:

To find the horizontal distance the ball travels before returning to its starting height, we can use the formula for the horizontal range of a projectile:

R = Vx * t

where R is the horizontal distance, Vx is the initial horizontal velocity, and t is the time of flight. Given that Vx = 33.9 m/s, we need to find the time of flight. Since the ball was launched at a 30-degree angle, we can use the formula:

t = 2 * Vy / g

where Vy is the initial vertical velocity and g is the acceleration due to gravity. Given that Vy = 19.6 m/s and g = 9.8 m/s², we can calculate t:

t = 2 * 19.6 m/s / 9.8 m/s² = 4 seconds

Now we can substitute the values into the formula for the horizontal range:

R = 33.9 m/s * 4 s = 135.6 meters

Therefore, the horizontal distance the ball travels before returning to its starting height is approximately 135.6 meters.